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+Central Limit Theorem
+Skip to main content
+
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 4.1 Central Limit Theorem
+
+
+
+
Section Title
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\(\newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
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+Section 4.1 Section Title
+
+
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+
\chi^2 Test
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 5.4 \(\chi^2\) Test
+
+
+
Conditional Probability
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 1.3 Conditional Probability
+
+
+Subsection 1.3.1 Conditional Probability
+
+
+Definition 1.3.1 .
+
+
+
Let \(A\) and \(B\) be events. The conditional probability of \(A\) given \(B\) is
+
+\begin{gather*}
+\Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)}.
+\end{gather*}
+
+
+
+Example 1.3.2 .
+
+
+
An experiment consists of rolling a fair 6-sided die two times. Let \(A\) be the event that the sum of the rolls is at least 10. To find \(\Pr(A)\text{,}\) we note that \(A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}\text{,}\) which has 6 elements. Since \(|\Omega| = 36\text{,}\) we have
+
+\begin{gather*}
+\Pr(A) = \frac{|A|}{|\Omega|} = \frac{6}{36} = \frac{1}{6}\text{.}
+\end{gather*}
+
+
+
+
Let \(B\) be the event that the first roll is 6. If we pause after the first die roll seeing the value of 6, we might be more inclined to expect a sum of at least 10. The evidence that we’ve already seen changes our understanding of the situation. Since \(|B| = 6\text{,}\) \(\Pr(B) = \frac{6}{36} = \frac{1}{6}\text{.}\) Also, \(A\cap B = \{(6, 4), (6, 5), (6, 6)\}\text{,}\) so \(\Pr(A\cap B) = \frac{3}{36} = \frac{1}{12}\text{.}\) Finally:
+
+\begin{gather*}
+\Pr(A \mid B) = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{1/12}{1/6} = \frac{1}{2}
+\end{gather*}
+
+
+
+Subsection 1.3.2 Diagnostic Testing
+
+Diagnostic tests for diseases aren’t perfect. When a test comes back positive or negative, a patient will want to understand the (conditional) probability that they have or don’t have the disease based on the evidence (the diagnostic test result).
+
+Definition 1.3.3 .
+
+The
sensitivity of a diagnostic test is the probability that a patient who has the disease will see a positive test result. The
specificity is the probability that a patient who does not have the disease will see a negative test result.
+
Introducing event notation, let
\(D\) be the event that a patient has the disease, and let
\(P\) be the event that they receive a positive test result. Then the sensitivity of the diagnostic test is
\(\Pr(P \mid D)\text{,}\) and the specificity is
\(\Pr(P^c \mid D^c)\text{.}\) However, when the patient takes a diagnostic test, the conditional probabilities they would be most interested in would be
\(\Pr(D \mid P)\) and
\(\Pr(D^c \mid P^c)\text{.}\)
+
Bayes’ Theorem expresses the relationship between a conditional probability
\(\Pr(A \mid B)\) and the flipped conditional probability
\(\Pr(B \mid A)\text{.}\)
+
+Theorem 1.3.4 . Bayes’ Theorem (v1).
+
+
+
+\begin{gather*}
+\Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A)}
+\end{gather*}
+
+
+
+
For example, if a patient sees a positive diagnostic test result, they might try to calculate:
+
+\begin{gather*}
+\Pr(D \mid P) = \frac{\Pr(P \mid D)\Pr(D)}{\Pr(P)}
+\end{gather*}
+
+
It will take some work to be able to use this formula. We likely don’t have direct access to \(\Pr(P)\) or \(\Pr(D)\text{.}\)
+
+
+
+
Let’s consider \(\Pr(P)\text{,}\) the probability of receiving a positive test result. The sensitivity \(\Pr(P \mid D)\) tells us this probability under the condition that the patient has the disease. For a patient who doesn’t have the disease, the specificity isn’t quite the number we’re looking for. However, consider the complementary probability:
+
+\begin{gather*}
+\Pr(P \mid D^c) = 1 - \Pr(P^c \mid D^c)
+\end{gather*}
+
+
The total \(\Pr(P)\) can be divided into two categories: patients who have the disease and test positive, and patients who don’t have the disease and test positive. So:
+
+\begin{align}
+\Pr(P) \amp = \Pr(P\cap D) + \Pr(P\cap D^c) \tag{1.3.1}\\
+\amp = \Pr(P\mid D)\Pr(D) + \Pr(P \mid D^c)\Pr(D^c) \tag{1.3.2}\\
+\amp = \Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c \mid D^c))\Pr(D^c) \tag{1.3.3}\\
+\amp = (\text{sensitivity})\Pr(D) + (1 - \text{ specificity})\Pr(D^c) \tag{1.3.4}
+\end{align}
+
+
We can take this breakdown of \(\Pr(P)\) and write a new version of Bayes’ Theorem:
+
+
+Theorem 1.3.5 . Bayes’ Theorem (v2).
+
+
+
+\begin{gather*}
+\Pr(B \mid A) = \frac{\Pr(A \mid B)\Pr(B)}{\Pr(A \mid B)\Pr(B) + (1 - \Pr(A^c \mid B^c))\Pr(B^c)}
+\end{gather*}
+
+
+
We’re still missing a crucial piece of information:
\(\Pr(D)\text{,}\) the probability (not conditioned on any evidence) that the patient has the disease. This is often referred to as the
prior , as in, our prior understanding of the probability of something before we gained some new information from evidence. The conditional probability calculated using Bayes’ Theorem is usually called the
posterior (i.e., after taking evidence into account). There isn’t always one single number that’s reasonable to use as the prior probability. For example, in a diagnostic testing situation, the
prevalence of the disease—i.e., the proportion of the population who have the disease—might feel like a natural number to use as the prior. However, what prevalence should you use? During the COVID-19 pandemic, the prevalence of COVID in a particular country, state, and city might be different. There’s also the possibility of applying Bayes’ Theorem multiple times to take into account multiple pieces of evidence, using the posterior probability from one application of Bayes’ Theorem to play the role of the prior probability in the next. This idea would apply if, for example, a patient took a second diagnostic test to double-check.
+
+Example 1.3.6 .
+
+A 50-year old woman with no symptoms is screened for breast cancer and tests positive. If the prevalence of breast cancer for women in her age group is 1% and the particular screening process used has a sensitivity of 90% and a specificity of 91%, what is the probability that the woman has breast cancer given her positive result?
+
Solution .
+
+
Let \(P\) be the event of testing positive and \(D\) the event of having the disease. Then the prevalence \(\Pr(D)\) is given as 1%, or 0.01. The sensitivity is \(\Pr(P\mid D) = 0.9\text{,}\) and the specificity is \(\Pr(P^c\mid D^c) = 0.91\text{.}\) So, according to Bayes’ Theorem:
+
+\begin{align*}
+\Pr(D\mid P) \amp = \frac{\Pr(P\mid D)\Pr(D)}{\Pr(P\mid D)\Pr(D) + (1 - \Pr(P^c\mid D^c))\Pr(D^c)} \\
+\amp = \frac{(0.9)(0.01)}{(0.9)(0.01) + (1 - 0.91)(1 - 0.01)} \\
+\amp \approx 0.092
+\end{align*}
+
+
+
+
+
+Exercises 1.3.3 Exercises
+
+
+
Exercise Group.
+
In each of the following scenarios with given events
\(A\) and
\(B\text{,}\) alculate
\(\Pr(A), \Pr(B)\text{,}\) \(\Pr(A\cap B)\text{,}\) \(\Pr(A \mid B)\text{,}\) and
\(\Pr(B \mid A)\text{.}\)
+
+
+
1. An experiment consists of rolling a fair die two times. Let
\(A\) be the event that the sum is even, and let
\(B\) be the event that the second roll is higher than the first.
+
2. An experiment consists of flipping a fair coin three times. Let
\(A\) be the event that the first and second flips match. Let
\(B\) be the event that there are at least two heads.
+
+
+
+
3. A diagnostic test is developed to detect a disease present in 3.2% of the population. For a patient who has the disease, the test will accurately give a positive result 65% of the time. When the patient does not have the disease, the test will accurately give a negative result 99.9% of the time.
+
(a) For a patient who receives a positive test, what is the probability they have the disease?
+
(b) For a patient who receives a negative test, what is the probability they do not have the disease?
+
+
Confidence Intervals
+Skip to main content
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 4.2 Confidence Intervals
+
+
+
+
Continuous Random Variables
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 2.2 Continuous Random Variables
+
+
+Subsection 2.2.1 Probability Density Functions
+
+We have to think about probability differently for continuous random variables. If
\(X\) is a continuous random variable taking values in an interval
\([a, b]\text{,}\) it doesn’t quite make sense to ask, as we have for discrete random variables, to find the probabilities of each individual outcome.
+
When we roll a fair 6-sided die, there are 6 equally likely outcomes, so they each get assigned the probability
\(1/6\text{.}\) If we try to select a random number in the interval
\([0, 1]\) in a similarly "fair" way (i.e., with each outcome equally likely), then what probability
\(\alpha\) should we assign to each outcome? If
\(\alpha > 0\text{,}\) then we have a problem: if you add up the probabilities of all the (infinitely many) outcomes, you get an infinite total, whereas the total probability in any sample space should be 1. If
\(\alpha = 0\text{,}\) then we also have a problem: if you add up the probabilities of all the outcomes, you get total of 0, whereas the total probability in any sample space should be 1.
+
We need a new way to model and calculate probabilities for continuous random variables.
+
+Definition 2.2.1 .
+
+
+
Let \(X\) be a continuous random variable taking values in the interval \([a, b]\text{.}\) The probability density function (or pdf ) for \(X\) is a function \(f(x)\) defined on \([a, b]\) satisfying the properties:
+
+
+
+
\(f(x) \geq 0\) for all
\(x \in [a, b]\text{,}\) and
+
+
+
+
\(\displaystyle{\int_a^b f(x)\ dx = 1}\text{.}\)
+
+
+
+
It’s important to keep in mind that the output values of a pdf
\(f(x)\) are
probability densities , not probabilities. You can (somewhat casually) think of the units of
\(f(x)\) values as "probability per <units of the sample space>". You may not have discussed the units of the differential
\(dx\) very much in your Calculus 1 course, but
\(dx\) has the same units as
\(x\) and is being multiplied by the integrand when you take an integral. So, although
\(f(x)\) is not a probability, the result of computing
\(\int_c^d f(x)\ dx\) is a probability.
+
+
A pdf’s purpose in life is to be integrated, which tells us probabilities for a continuous random variable:
+
+\begin{gather*}
+\Pr(c \leq X \leq d) = \int_c^d f(x)\ dx.
+\end{gather*}
+
+
+
+Definition 2.2.2 .
+
+
+
+
We’ll ignore for the moment how to find this pdf and instead we’ll quickly check that it behaves the way we want it to:
+
+\begin{align*}
+\Pr(c \leq X \leq d) \amp = \int_c^d \frac{1}{b-a}\ dx \\
+\amp = \frac{x}{b - a}\bigg|_c^d \\
+\amp = \frac{d - c}{b - a},
+\end{align*}
+
+
as we wanted.
+
+
+Example 2.2.3 .
+
+
+
Let \(X\) be a continuous random variable taking values in the interval \([0, 4]\) with pdf \(f(x) = k \sqrt{x}\) for some constant \(k\text{.}\) Because of the restriction that the total probability must be 1, we can determine the value of \(k\text{:}\)
+
+
+\begin{align*}
+1 = \Pr(0\leq X \leq 4) \amp = \int_0^4 k\sqrt{x}\ dx \\
+\amp = \frac{kx^{3/2}}{3/2}\bigg|_0^4 \\
+\amp = \frac{2kx^{3/2}}{3}\bigg|_0^4 \\
+\amp = \left(\frac{2k(4^{3/2})}{3}\right) - 0 \\
+\amp = \frac{16k}{3}
+\end{align*}
+
+
Therefore \(k = \frac{3}{16}\text{.}\) Now we can calculate probabilities, e.g.:
+
+\begin{align*}
+\Pr(1 \leq X \leq 2) \amp = \int_1^2 \frac{3\sqrt{x}}{16}\ dx \\
+\amp = \frac{3}{16}\left(\frac{2x^{3/2}}{3}\right)\bigg|_1^2 \\
+\amp = \frac{1}{8}\left(x^{3/2}\right)\bigg|_1^2 \\
+\amp = \frac{1}{8}\left( 2^{3/2} - 1 \right) \\
+\amp \approx 0.23.
+\end{align*}
+
+
+
+Subsection 2.2.2 Cumulative Distribution Functions
+
+When a random variable takes values in an interval
\([a, b]\text{,}\) it’s often useful to consider how much probability has been accumulated up to some specified value
\(x \in [a, b]\text{.}\) For each value of
\(x\text{,}\) we can calculate this accumulated probability. Since there’s an association between inputs (the
\(x\) values) and output (the accumulated probability), we can naturally build a function.
+
+Definition 2.2.4 .
+
+
+
Let \(X\) be a continuous random variable taking values in an interval \([a, b]\text{.}\) The cumulative distribution function (or cdf ) is the function \(F(x)\) defined on \([a, b]\) such that
+
+\begin{gather*}
+F(x) = \Pr(a \leq X \leq x).
+\end{gather*}
+
+
+
+
How do we actually calculate these cumulative probabilities? Luckily, these are precisely the sort of probabilities we can use the pdf to find:
+
+\begin{gather*}
+F(x) = \Pr(a \leq X \leq x) = \int_a^x f(t)\ dt.
+\end{gather*}
+
+
(Note that on the far left side of the equation, we used the letter \(x\) to refer to some point within \([a, b]\) up to which we’re trying to accumulate probability. So that value \(x\) must also appear at the top of the integral sign. We can’t then use the same letter for the integrand and differential inside the integral; we can’t use the same letter to refer to two different variables at the same time. So, we have changed the letter inside the integral to \(t\text{.}\) )
+
+
+
+
+\begin{gather*}
+F'(x) = f(x).
+\end{gather*}
+
+
Let’s see how we can put this relationship to use.
+
+
+Example 2.2.5 .
+
+
+
Suppose
\(X\) is uniformly distributed in the interval
\([a, b]\text{.}\) We previously said in
Definition 2.2.2 that, for any subinterval
\([c, d] \subset [a, b]\text{,}\) we should have
+
+\begin{gather*}
+\Pr(c\leq X \leq d) = \frac{d - c}{b - a}.
+\end{gather*}
+
+
In particular, for any \(x\in [a, b]\text{,}\) we would have:
+
+\begin{gather*}
+F(x) = \Pr(a \leq X \leq x) = \frac{x - a}{b - a}.
+\end{gather*}
+
+
Therefore, taking the derivative:
+
+\begin{gather*}
+f(x) = F'(x) = \frac{1}{b - a}.
+\end{gather*}
+
+
+
So, we can see that the idea we wanted to capture in probability led to a formula that gave accumulated probability, which then led to the pdf. Both the cdf and the pdf contain essentially all of the information we need to answer probability questions about
\(X\text{.}\) However, keep in mind that these functions output different mathematical objects.
\(F(x)\) outputs probabilities already;
\(f(x)\) must be integrated over a subinterval to find probabilities.
+
+Subsection 2.2.3 The Exponential Distribution
+
+
+Definition 2.2.6 .
+
+Suppose we observe a Poisson process in which an event occurs with rate parameter
\(\lambda\text{.}\) Let
\(T\) be the time until the event is first observed. Then
\(T\) has the
exponential distribution with parameter
\(\lambda\text{.}\) We’ll write
\(T \sim \Exp(\lambda)\) to denote this.
+
+Fact 2.2.7 . Exponential Density Function.
+
+
+
+\begin{gather*}
+f(t) = \lambda e^{-\lambda t}
+\end{gather*}
+
+
+
+
The binomial, geometric, Poisson, and exponential distributions are conceptually grouped together. Consider the following two dichotomies:
+
+
+Is time being measured discretely or continuously?
+
+
+Does the random variable count occurrences of an event, or does it measure time until an event?
+
+
+
There are four combinations of answers to these two questions.
+
+Binomial
+
+"Time" is discrete; the random variables counts occurrences.
+
Geometric
+
+"Time" is discrete; the random variable measures "time" until an event.
+
Poisson
+
+Time is continuous; the random variable counts occurrences.
+
Exponential
+
+Time is continuous; the random variable measures time until an event.
+
+
When we talk about discrete "time", we don’t always mean something measured in actual units of time. We might; for example, an experiment may involve a bacterial colony growing in a petri dish over time, but we only check in and measure something about the bacterial colony once each day. In this case, since our measurements of time must occur in whole numbers of days, time is being measured in a discrete way (that is, not involving an interval of real numbers).
+
+
Consider, instead, an experiment in which we flip a coin repeatedly until we first see heads appear. The number of flips doesn’t have units of time, but counting flips is analogous to measuring time until an event occurs. The flip count is a discrete stand-in for something analogous to time. We can consider this experiment to be a discrete analog of a Poisson process.
+
It’s also worth including a warning here about the Poisson distribution, specifically. The dichotomy mentioned above about whether time is being measured discretely or continuously is
not the same as the consideration of whether the random variable itself is discrete or continuous. In a Poisson process, time is being measured continuously. But the Poisson random variable counts the number of occurrences of the event, and this number must be a whole number. So a Poisson random variable is a discrete random variable.
+
+Subsection 2.2.4 Normal Distribution
+
+
+Definition 2.2.8 .
+
+
+
Consider the parametrized family of functions:
+
+\begin{gather*}
+f(x; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x-\mu)^2}{2\sigma^2}}.
+\end{gather*}
+
+
The function \(f(x; \mu, \sigma^2)\) is called a normal density function with parameters \(\mu\) and \(\sigma^2\text{.}\) A continuous random variable \(X\) with pdf \(f(x; \mu, \sigma^2)\) is said to have a normal distribution . In the specific case that \(\mu = 0\) and \(\sigma^2 = 1\text{,}\) we call the resulting distribution the standard normal distribution .
+
+
It’s an unfortunate reality that the normal density functions can’t easily be integrated. (In fact, it is provably impossible to write nice formulas for their antiderivatives). So we won’t worry much about probability computations like
\(\Pr(c \leq X \leq d)\) for normal distributions. The normal distributions will play a
central role in later chapters (after which you can return here and appreciate this pun).
+
+Exercises 2.2.5 Exercises
+
+1. A continuous random variable
\(X\) taking values in
\([1, 4]\) has p.d.f.
\(f(x) = k(x - \sqrt{x})\) for some constant
\(k\text{.}\) What is the value of
\(k\text{?}\)
+
2. A continuous random variable
\(X\) taking values in
\([1, 2]\) has p.d.f.
\(\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}\text{.}\) Find the c.d.f.
\(F(x)\text{.}\) Use your c.d.f. to find
\(\Pr\left(1 \leq X \leq \frac{3}{2}\right)\text{.}\)
+
3. A continuous random variable
\(X\) taking values in
\([2, 3]\) has c.d.f.
\(F(x) = \frac{x^3}{3} - x^2 + 4\text{.}\) Find the p.d.f.
\(f(x)\text{.}\)
+
4. A radioactive material emits particles at an unknown probabilistic rate
\(\lambda\) particles per minute. We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 minutes. Write the likelihood function
\(\mathcal{L}(\lambda)\) based on this data. What is the maximum likelihood estimation for
\(\lambda\text{?}\)
+
5. Suppose a parameter
\(\theta\) takes values in
\([0, 1]\) with likelihood function
\(\mathcal{L}(\theta) = \sqrt{\theta} - \theta^2\text{.}\) Find the maximum likelihood estimation of
\(\theta\text{.}\)
+
+
Correlation
+Skip to main content
+
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 6.1 Correlation
+
+
+
+
Covariance
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 3.3 Covariance
+
+
+
+Exercises Exercises
+
+1.
+
Discrete Random Variables
+Skip to main content
+
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 2.1 Discrete Random Variables
+
+
+Subsection 2.1.1 Discrete Random Variables
+
+
+Definition 2.1.1 .
+
+Let
\(\Omega\) be a sample space. A
random variable is a function
\(X \colon \Omega \to \R\) which assigns a number to each outcome.
+
+Example 2.1.2 .
+
+Suppose an experiment consists of flipping a fair coin 4 times. Let
\(X\) be the number of heads. The actual outcomes in the experiment are heads/tails sequences of length 4, such as
\(HTHT\) and
\(HHHT\text{.}\) The random variable
\(X\) assigns a numerical measurement to the outcomes, such as
\(X(HTHT) = 2\) and
\(X(HHHT) = 3\text{.}\)
+
For any real number value
\(a\text{,}\) we can consider the event consisting of outcomes such that
\(X = a\text{.}\) For example, the flip sequences
\(HHHT, HHTH, HTHH\text{,}\) and
\(THHH\) all have 3 heads, so
\(\Pr(X = 3) = \frac{4}{16} = \frac{1}{4}\text{.}\)
+
+Example 2.1.3 . Indicator Random Variable.
+
+
+
Let \(A\) be an event in the sample space \(\Omega\text{.}\) An indicator random variable for \(A\) is the random variable \(X\) such that
+
+\begin{align*}
+X(\omega) = \begin{cases} 1 \amp \omega \in A \\ 0 \amp \omega \notin A \end{cases}
+\end{align*}
+
+
+
Indicator random variables are also called
Bernoulli random variables , although we’ll prefer the former term since it more clearly states the purpose of these random variables: to indicate whether or not a particular event has occurred. We’ll use the language "
\(X\) indicates
\(A\) " to mean that
\(X\) is an indicator random variable for the event
\(A\text{.}\) In this case,
\(\Pr(X = 1) = \Pr(A)\text{,}\) and
\(\Pr(X = 0) = 1 - \Pr(A)\text{.}\)
+
+Definition 2.1.4 .
+
+Consider the set
\(S\) of all real numbers which are actually output by a random variable
\(X\text{.}\) If
\(S\) does not contain any interval of values, then the random variable
\(X\) is called
discrete . Otherwise, it’s called
continuous .
+
+Example 2.1.5 .
+
+Any random variable
\(X\) defined on a finite sample space
\(\Omega\) is discrete—the set of outputs of
\(X\) cannot contain an interval if it only has finitely many values.
+
Suppose an experiment consists of growing a plant in a new fertilizer, and a random variable
\(Y\) measures the height of the plant after a set growing period. Now
\(Y\) could conceivably take on an interval’s worth of values (such as, e.g., any real number between 10 inches and 20 inches), so this random variable would be continuous.
+
The distinction between discrete and continuous random variables is not simply the distinction of whether the sample space is finite or infinite. Consider the experiment in which we flip a coin repeatedly until we first see a coin come up heads. Let
\(Z\) be the number of times the coin is flipped. Then there are infinitely many possible values of
\(Z\) (1, 2, 3, and so on), but there’s no interval of real numbers which are all possible outputs of
\(Z\text{.}\) The possible outputs of
\(Z\) are discrete (in the non-technical, English sense of the word: separated).
+
There are several particular types of distributions that appear repeatedly, since they model common behaviors. Indicator random variables, for example, will appear any time a random process generates a yes/no or success/failure type of answer to a question. Below are some other common distributions.
+
+Subsection 2.1.2 The Binomial Distribution
+
+
+Definition 2.1.6 .
+
+Suppose an event occurs with probability
\(p\text{.}\) If we perform
\(n\) independent trials, let
\(S\) be the random variable which counts the number of trials in which the event occurred. Then
\(S\) has the
binomial distribution with parameters
\(n\) and
\(p\text{.}\) We’ll write
\(S \sim \Bin(n, p)\) to denote this. For each value
\(0 \leq k \leq n\text{,}\) we’ll write
\(b(k)\) for
\(\Pr(S = k)\text{.}\) (If we want to keep track of the parameter values, we may write
\(b(k; n, p)\text{.}\) )
+
+Example 2.1.7 .
+
+Suppose we have a coin which comes up heads with probability
\(p = 0.4\text{.}\) If we flip the coin three times and let
\(S\) be the number of heads, then
\(S\) is binomially distributed with parameters
\(n = 3\) and
\(p = 0.4\text{.}\) Then
\(S\) has the distribution below.
+
Table 2.1.8. \(\Bin(3, 0.4)\)
+
+
+0
+0.216
+
+
+1
+0.432
+
+
+2
+0.288
+
+
+3
+0.064
+
+
+
To understand how these probabilities are calculated, consider a particular heads/tails sequence, such as \(HHHT\text{.}\) If the probability of heads is 0.4, then the probability of tails is 0.6. Since the results of different coin flips are independent from each other, the probability of seeing precisely the sequence \(HHT\) would be the product \((0.4)(0.4)(0.6) = 0.096\text{.}\) The probability of seeing precisely \(HTH\) would be \((0.4)(0.6)(0.4) = 0.096\text{,}\) the same product of terms in a different order. So, if we knew exactly the number of flip sequences of length 3 with exactly 2 heads, we could calculate the probability of seeing 2 heads as:
+
+\begin{gather*}
+\Pr(S = 2) = (\text{number of sequences})(\text{probability of each sequence})
+\end{gather*}
+
+
The probability for each sequence is straightforward to generalize: if there are \(n\) flips and \(k\) of them are heads, then \(n - k\) of them are tails. So the product we should calculate will have \(k\) copies of the parameter \(p\) and \(n - k\) copies of the complementary probability \(1 - p\text{:}\)
+
+
+\begin{gather*}
+\text{probability of each sequence } = p^k(1-p)^k
+\end{gather*}
+
+
+
+Definition 2.1.9 .
+
+We’ll write the symbol
\({n \choose k}\) to mean the number of ways to choose
\(k\) things out of
\(n\) things. The number
\({n \choose k}\) is read "
\(n\) choose
\(k\) " and called a
binomial coefficient .
+
The term "binomial coefficient" is used here because these numbers form coefficients of a polynomial in an important theorem about binomials called the
Binomial Theorem . The binomial coefficients will precisely tell us the number of flip sequences of length
\(n\) with precisely
\(k\) heads (to build such a sequence, we must choose
\(k\) out of
\(n\) flips in the sequence to be heads; the rest must be tails). So, now we need a way to find the numbers
\({n \choose k}\text{.}\)
+
+Definition 2.1.10 .
+
+
+
Let \(n\) be a positive integer. Then the factorial of \(n\text{,}\) written \(n!\text{,}\) is the product of the positive integers up to \(n\text{,}\) i.e.:
+
+\begin{gather*}
+n! = 1 \times 2 \times 3 \times \dotsb \times (n-1) \times n.
+\end{gather*}
+
+
+
+Fact 2.1.11 . Binomial Coefficient Formula.
+
+
+
+
We won’t provide a justification of this formula here, but we’ll put it to use to calculate probabilities in the binomial distribution. For example, the number of flip sequences of length 3 with exactly 2 heads is:
+
+\begin{gather*}
+{3 \choose 2} = \frac{3!}{2!1!} = \frac{6}{2} = 3.
+\end{gather*}
+
+
Since we previously said the probability of such a sequence is 0.096, we can now find the probability of seeing exactly 2 heads in 3 flips of the coin:
+
+\begin{align*}
+\Pr(S = 2) \amp = (\text{number of sequences})(\text{probability of each sequence}) \\
+\amp = (3)(0.096) \\
+\amp = 0.288,
+\end{align*}
+
+
+
+
+Fact 2.1.12 . Binomial Probability Formula.
+
+
+
+Subsection 2.1.3 Geometric Distribution
+
+
+Definition 2.1.13 .
+
+Suppose an event occurs with probability
\(p\text{.}\) If we perform independent trials repeatedly, let
\(N\) be the random variable which counts the number of trials performed until we see the event occur for the first time. (
\(N\) counts the final trial in which the event occurs.) Then
\(N\) has the
geometric distribution with parameter
\(p\text{.}\) We’ll write
\(N \sim \Geom(p)\) to denote this. For each value
\(k \geq 1\text{,}\) we’ll write
\(g(k)\) for
\(\Pr(N = k)\text{.}\) (If we want to keep track of the parameter value, we may write
\(g(k; p)\text{.}\) )
+
Suppose we flip a coin repeatedly until we first see heads. The number
\(N\) of flips is geometrically distributed with parameter
\(p\text{,}\) which gives the probability of a flip coming up heads. Then the probability of tails is
\(1 - p\text{.}\) For any particular
\(k \geq 1\text{,}\) in order to have
\(N = k\text{,}\) we must see exactly the flip sequence
\(TT\dotsm TH\text{,}\) where the number of tails is exactly
\(k - 1\text{.}\) The probability of seeing this particular sequence is then:
+
+Fact 2.1.14 . Geometric Probability Formula.
+
+
+
+Subsection 2.1.4 Poisson Distribution
+
+
+Definition 2.1.15 .
+
+A
Poisson process is one in which some event occurs randomly at a constant probabilistic rate
\(\lambda\) over time. Suppose we observe a Poisson process for a fixed amount of time
\(t\text{,}\) and let
\(N\) count the number of occurrences of the event. Then
\(N\) has the
Poisson distribution with parameters
\(\lambda\) and
\(t\text{.}\) We’ll write
\(N \sim \Poiss(\lambda, t)\) to denote this. For each value
\(k \geq 0\text{,}\) we’ll write
\(p(k)\) for
\(\Pr(N = k)\text{.}\) (If we want to keep track of the parameter values, we may write
\(p(k; \lambda, t)\text{.}\) See
Example 2.1.17 for an important detail about units.)
+
+Fact 2.1.16 . Poisson Probability Formula.
+
+
+The term
\(\lambda t\) is potentially misleading here depending on the units given for each value. The idea is that the rate information might be given for a different period of time than the observation period, so the rate should be scaled before calculating a probability. But the rate and observation time must be using the same time unit before multiplying. See the following example.
+
+Example 2.1.17 .
+
+
+
Suppose we observe traffic along a particular stretch of highway which typically has 200 cars pass per hour. Let \(N\) be the number of cars seen in a 2-hour observation period. Since the parameter \(\lambda = 200\) cars per hour is given for a 1-hour period, we should scale it to find the rate for a 2-hour period: \(\lambda t = (200)(2) = 400\) cars per 2-hours. Then, for example:
+
+\begin{gather*}
+\Pr(N = 375) = p(375; 200, 2) = \frac{400^{375}}{375!} e^{-400} \approx 0.01.
+\end{gather*}
+
+
+
+
Now consider the random variable \(M\) counting the number of cars seen in a 30-minute observation period. We should not calculate \(\lambda t = (200)(30) = 6000\text{,}\) since \(\lambda\) measures time in hours and \(t\) measures time in minutes. We should first convert \(t = 0.5\) hours, then we can find our appropriately scaled rate information: \((200)(0.5) = 100\) cars per half-hour. Then, for example:
+
+\begin{gather*}
+\Pr(M = 110) = p(110; 200, 0.5) = \frac{100^{110}}{110!}e^{-100} \approx 0.02.
+\end{gather*}
+
+
Note that when we write the notation \(p(k; \lambda, t)\text{,}\) we assume that \(\lambda\) and \(t\) are expressed with the same time units already.
+
+
+Exercises 2.1.5 Exercises
+
+1. We flip a coin with unknown bias 100 times and observe 43 heads. What is the maximum likelihood estimation for the probability of the coin coming up heads?
+
2. An experiment consists of flipping a biased coin 20 times. If the coin comes up heads with probability
\(p = 0.3\text{,}\) find the probability of seeing 5 heads. Find the probability of seeing up to (and including) 3 heads.
+
3. An experiment consists of flipping a coin repeatedly until we first see heads.
+
4. If the coin comes up heads with probability 0.4, what is the probability we’ll see our first heads within three flips? What about precisely on the third flip?
+
5. Suppose the coin has an unknown probability of coming up heads. We perform the experiment in five independent trials, during which it takes 4, 5, 4, 3, and 6 flips to see our first heads in each trial. What is the maximum likelihood estimation for the probability of the coin coming up heads on a flip?
+
6. A particular store has an average of 20 customers each hour. During a 4-hour afternoon shift, what is the probability of serving 80 customers.
+
7. A particular store owner wants to approximate the average hourly rate at which customers come into the store. They observe 80 customers enter during a particular 4-hour shift. What is the maximum likelihood estimation for the hourly customer rate?
+
+
Section Title
+Skip to main content
+
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\(\newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 2.1 Section Title
+
+
+
+
Expected Value
+Skip to main content
+
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 3.1 Expected Value
+
+
+Subsection 3.1.1 Discrete Expected Value
+
+A full probability distribution (for a discrete random variable) or a density function (for a continuous random variable) carry all of the probability information. Often, we seek
statistics , i.e., numbers that summarize a distribution in some way.
+
+Definition 3.1.1 .
+
+
+
Let \(X\) be a discrete random variable taking the values \(x_1, x_2, \dotsc, x_n\text{.}\) The expected value (also called mean , or expectation ) of \(X\) is the weighted average of the values of \(X\text{,}\) where the weights are the probabilities of \(X\) taking each value:
+
+\begin{gather*}
+\E(X) = \sum_{i=1}^n x_i \Pr(X = x_i).
+\end{gather*}
+
+
+
+Example 3.1.2 .
+
+
+
If \(X\) takes on the values \(x_1, x_2, \dotsc, x_n\) uniformly (i.e., each value having probability \(\frac{1}{n}\) ), then the expected value is the usual average:
+
+\begin{align*}
+\E(X) \amp = x_1 \cdot \frac{1}{n} + x_2\cdot \frac{1}{n} + \dotsb + x_n \cdot \frac{1}{n} \\
+\amp = \frac{x_1 + x_2 + \dotsb + x_n}{n}
+\end{align*}
+
+
For example, the expected value of a fair 6-sided die roll \(R\) is:
+
+\begin{gather*}
+\E(R) = \frac{1 + 2 + \dotsb + 6}{6} = \frac{21}{6} = \frac{7}{2}.
+\end{gather*}
+
+
+
+Example 3.1.3 .
+
+Suppose we roll a fair 6-sided die two times, and let
\(S\) be the sum of the rolls. It’s straightforward to check the distribution for
\(S\) shown below. We’ve also included a column containing the products
\(k \Pr(S = k)\text{,}\) which must be summed up to find
\(\E(S)\text{.}\)
+
Table 3.1.4.
+
+
+2
+\(1/36\) \(2/36\)
+
+3
+\(2/36\) \(6/36\)
+
+4
+\(3/36\) \(12/36\)
+
+5
+\(4/36\) \(20/36\)
+
+6
+\(5/36\) \(30/36\)
+
+7
+\(6/36\) \(42/36\)
+
+8
+\(5/36\) \(40/36\)
+
+9
+\(4/36\) \(36/36\)
+
+10
+\(3/36\) \(30/36\)
+
+11
+\(2/36\) \(22/36\)
+
+12
+\(1/36\) \(12/36\)
+
The sum of the third column gives
\(\E(S) = \frac{252}{36} = 7\text{.}\)
+
+Example 3.1.5 .
+
+
+
Let \(X\) indicate an event \(A\) which has probability \(p\text{.}\) Then:
+
+\begin{gather*}
+\E(X) = 0 \cdot (1-p) + 1 \cdot p = p.
+\end{gather*}
+
+
That is, the expected value of an indicator random variable is the probability of the event that it indicates.
+
+
Having found the expected value for one named distribution, we may be tempted to run through our other discrete distributions (binomial, geometric, and Poisson) and find nice formulas for their expected values. Unfortunately, both the geometric distribution and the Poisson distribution have infinitely many possible values, and the summation of infinitely many discrete terms is beyond the scope of this course. (A Calculus 2 course covering series would give the right tools to find expected value formulas with full justification for these distributions.)
+
The binomial distribution takes only finitely many values, and we have
Fact 2.1.12 that lets us generate a distribution table reasonably quickly, given the parameters
\(n\) and
\(p\text{.}\) Feel free to pick some parameter values (keep
\(n\) fairly small!) to try this calculation by hand, but we’ll use a powerful theorem a bit later to give a very nice general formula depending on
\(n\) and
\(p\text{.}\)
+
+Subsection 3.1.2 Continuous Expected Value
+
+
+
+Subsection 3.1.3 Linearity of Expected Value
+
+
+
+Exercises 3.1.4 Exercises
+
+1. Consider a random variable
\(X\) with probability distribution below. Find
\(\E(X)\text{.}\)
+
Table 3.1.6.
+
+
+1
+0.1
+
+
+2
+0.05
+
+
+3
+0.2
+
+
+4
+0.15
+
+
+5
+0.15
+
+
+6
+0.1
+
+
+7
+0.05
+
+
+8
+0.1
+
+
+9
+0.1
+
+
2. Let
\(X\) and
\(Y\) be random variables each taking the values 1, 2, 3, 4, 5. Write different distribution tables for
\(X\) and
\(Y\) so that they have the same expected value.
+
3. Suppose we flip a coin
\(n = 100\) times, and let
\(N\) count the number of heads. If the coin comes up heads on a flip with probability
\(p = 0.4\text{,}\) what is
\(\E(N)\text{?}\) What if
\(n = 80\) and
\(p = 0.6\text{?}\) What if
\(n = 200\) and
\(p = 0.5\text{?}\)
+
4. If
\(\E(X) = 3\text{,}\) \(\E(Y) = -2\text{,}\) and
\(\E(Z) = 1\text{,}\) what is
\(\E(4X + 5Y - Z + 3)\text{?}\)
+
5. A continuous random variable
\(X\) taking values in
\([0, 1]\) has p.d.f.
\(f(x) = 2x\text{.}\) What is
\(\E(X)\text{?}\)
+
6. A continuous random variable
\(X\) taking values in
\([-1, 1]\) has p.d.f.
\(f(x) = \frac{3x^2}{2}\text{.}\) What is
\(\E(X)\text{?}\)
+
7. A continuous random variable
\(X\) taking values in
\([1, 4]\) has p.d.f.
\(f(x) = \frac{4}{3x^2}\text{.}\) What is
\(\E(X)\text{?}\)
+
8. A continuous random variable
\(X\) taking values in
\([1, 4]\) has p.d.f.
\(f(x) = k(x - \sqrt{x})\) for some constant
\(k\text{.}\) In a previous problem, you found the value of
\(k\text{.}\) Now, find
\(\E(X)\text{.}\)
+
9. A continuous random variable
\(X\) taking values in
\([1, 2]\) has p.d.f.
\(\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}\text{.}\) Find
\(\E(X)\text{.}\)
+
+
Independent Events
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 1.4 Independent Events
+
+The idea of conditional probability is that knowledge of one event can change our understanding of the probability of another. But it’s also important to understand when this is not the case.
+
+Definition 1.4.1 .
+
+Events
\(A, B\) are
independent if
\(\Pr(A \mid B) = \Pr(A)\text{.}\)
+
+
The equation \(\Pr(A\mid B) = \Pr(A)\) very concretely says: knowledge that the event \(B\) has occurred does not change our understanding of \(\Pr(A)\text{.}\) Assuming \(\Pr(A) \neq 0\text{,}\) this is equivalent to \(\Pr(A\cap B) = \Pr(A)\Pr(B)\text{,}\) since:
+
+\begin{align*}
+\Pr(A\mid B) \amp = \Pr(A) \\
+\frac{\Pr(A\cap B)}{\Pr(B)} \amp = \Pr(A) \\
+\Pr(A\cap B) \amp = \Pr(A)\Pr(B)
+\end{align*}
+
+
This latter form of the equation is often more useful in calculations.
+
+
+Example 1.4.2 .
+
+In
Example 1.3.2 , we considered an experiment in which we rolled a fair 6-sided die twice. We defined events
\(A\) that the sum of the rolls is at least 10 and
\(B\) that the first roll is a 6, and we found that
\(\Pr(A \mid B) \neq \Pr(A)\text{,}\) so
\(A\) and
\(B\) are not independent.
+
+
Now consider the event \(C\) that the sum of the rolls is 7, which sounds very similar to the event \(A\text{.}\) We have:
+
+\begin{align*}
+C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} \\
+\Pr(C) \amp = \frac{6}{36} = \frac{1}{6} \\
+B\cap C \amp = \{(6, 1)\} \\
+\Pr(B\cap C) \amp = \frac{1}{36} \\
+\text{therefore: } \Pr(C\mid B) \amp = \frac{\Pr(C\cap B)}{\Pr(B)} = \frac{1/36}{1/6} = \frac{1}{6} = \Pr(C)
+\end{align*}
+
+
So, even though the descriptions of events \(A\) and \(C\) are very similar, the event \(C\) is independent with \(B\text{,}\) while \(A\) is not.
+
+
+Exercises Exercises
+
+1. An experiment consists of rolling a fair die two times. Let
\(A\) be the event that the sum is even, and let
\(B\) be the event that the second roll is higher than the first. Are
\(A\) and
\(B\) independent?
+
2. An experiment consists of flipping a fair coin three times. Let
\(A\) be the event that the first and second flips match. Let
\(B\) be the event that there are at least two heads. Are
\(A\) and
\(B\) independent?
+
3. Let
\(A = \{1, 2, 3\}\) and
\(B = \{3, 4, 5\}\) be events in the sample space
\(\Omega = \{1, 2, 3, 4, 5, 6\}\text{.}\) Create a probability distribution for
\(\Omega\) so that
\(A, B\) are independent.
+
+
Linear Regression
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 6.2 Linear Regression
+
+
+
+
One Sample Tests
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Contents
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+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 5.1 One Sample Tests
+
+
+
+
Power of a Test
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+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 5.3 Power of a Test
+
+
+
+
Definition of Probability
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\(\newcommand{\N}{\mathbb N}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 1.2 Definition of Probability
+
+
+Definition 1.2.1 .
+
+
+
A probability distribution on a sample space \(\Omega\) assigns probabilities to every event, satisfying the following conditions:
+
+
+
+
\(\Pr(\Omega) = 1\text{.}\)
+
+
+
+
\(0 \leq \Pr(A) \leq 1\) for any event
\(A\text{.}\)
+
+
+If
\(A \cap B = \emptyset\text{,}\) then
\(\Pr(A\cup B) = \Pr(A) + \Pr(B)\text{.}\)
+
+
+
+
For small probability spaces (i.e., with finitely many outcomes in the sample space), we’ll usually assign probabilities to each individual outcome, and perhaps list them in a table. Then, to find the probability of any event, simply add together the probabilities of each outcome in that event.
+
+Example 1.2.2 .
+
+An experiment consists of rolling a standard 6-sided die. The sample space is
\(\Omega = \{1, 2, 3, 4, 5, 6\}\text{.}\) The probability distribution (assuming a fair die) is shown below.
+
Table 1.2.3. Distribution for a fair die
+
+\(x\) \(\Pr(x)\)
+
+1
+\(1/6\)
+
+2
+\(1/6\)
+
+3
+\(1/6\)
+
+4
+\(1/6\)
+
+5
+\(1/6\)
+
+6
+\(1/6\)
+
+
One possible event is \(A = \{2, 4, 6\}\text{,}\) i.e., the event that the result of the roll is even. The probability of \(A\) is:
+
+\begin{gather*}
+\Pr(A) = \Pr(2) + \Pr(4) + \Pr(6) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}
+\end{gather*}
+
+
+
+Definition 1.2.4 .
+
+
+
+Example 1.2.5 .
+
+A person picks a random number from 1 to 10. What is the probability that they picked 3?
+
+Exercises Exercises
+
+1. Consider the sample space
\(\Omega = \{1, 2, 3, 4, 5, 6, 7, 8\}\) with probability distribution below. Calculate the probabilities of
\(A = \{1, 3, 7, 8\}\text{,}\) \(B = \{2, 3, 6, 7\}\text{,}\) \(A\cup B\text{,}\) and
\(A \cap B\text{.}\)
+
Table 1.2.6.
+
+\(x\) \(\Pr(x)\)
+
+1
+0.1
+
+
+2
+0.05
+
+
+3
+0.2
+
+
+4
+0.15
+
+
+5
+0.15
+
+
+6
+0.1
+
+
+7
+0.05
+
+
+8
+0.1
+
+
+9
+0.1
+
+
2. Suppose we flip a coin two times. Answer the questions below. What about three flips? What about four flips?
+
(a) Write all outcomes in the sample space
\(\Omega\text{.}\)
+
(b) Make a probability distribution table for
\(\Omega\) assuming the coin is fair.
+
(c) Make a probability distribution table assuming the coin comes up heads with probability 0.3.
+
3. Suppose we roll a die two times. Answer the questions below.
+
(a) Write all outcomes in the sample space
\(\Omega\text{.}\)
+
(b) Make a probability distribution table for
\(\Omega\) assuming the die is fair.
+
(c) Let
\(A\) be the event that the second roll is higher than the first, and let
\(B\) be the event that the first roll is even. Find
\(\Pr(A), \Pr(B)\text{,}\) and
\(\Pr(A \mid B)\text{.}\)
+
4. Suppose a die has the values
\(1, 2, 3, 4, 5, 6\) on the faces, but the die is not fair. Instead, the probabilities scale by the same amount as the face values. For example, a result of 4 is twice as likely as a result of 2, since 4 is twice as large as 2; a result of 6 is six times more likely than a result of 1; and so on. Write a probability distribution table for this die.
+
5. Suppose a die has the values
\(1, 2, 3, 4, 5, 6\) on the faces, but the die is not fair. Instead, each even value has an equal probability, each odd value has an equal probability, and the even values are each twice as likely as the odd values to appear on a roll. Write a probability distribution table for this die.
+
6. A toxin molecule inside a cell has a 0.3 probability of leaving the cell during a 1-minute period. For each value of
\(n = 1, 2, 3, \dotsc\text{,}\) find the probability of the toxin molecule leaving the cell during the
\(n\) th minute. What is the probability of the molecule leaving the cell during the first 3 minutes?
+
7. Each of 10 toxin molecules inside a cell has a 0.3 probability of leaving the cell during a 1-minute period. For each value of
\(n = 1, 2, 3, \dotsc\text{,}\) and for each value of
\(0\leq k \leq n\text{,}\) find the probability that exactly
\(k\) toxin molecules remain in the cell after the
\(n\) th minute.
+
+
Set Theory
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+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 1.1 Set Theory
+
+
+Definition 1.1.1 .
+
+The
sample space , often denoted
\(\Omega\text{,}\) is the set of all possible results of an experiment. A single result is called an
outcome , while a collection of results is called an
event .
+
+Example 1.1.2 .
+
+An experiment consists of rolling a standard 6-sided die. The sample space is
\(\Omega = \{1, 2, 3, 4, 5, 6\}\text{.}\) One possible event is
\(A = \{2, 4, 6\}\text{,}\) i.e., the event that the result of the roll is even.
+
What would the sample space look like if we roll the die two times and recorded the results?
+
Answer .
+
+
+\begin{align*}
+\Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \\
+\amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \\
+\amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \\
+\amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \\
+\amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \\
+\amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}
+\end{align*}
+
+
+
+
+
+Definition 1.1.3 .
+
+Let
\(A\) be a set. The symbol
\(\in\) means "is an element of", as in
\(a \in A\text{.}\) Given another set
\(B\text{,}\) we say
\(A\) is a
subset of
\(B\text{,}\) written
\(A\subset B\text{,}\) to mean that every element of the set
\(A\) is also an element of the set
\(B\text{.}\)
+
+Definition 1.1.4 .
+
+
+
Consider sets \(A\) and \(B\text{,}\) each contained inside \(\Omega\text{.}\) We can combine sets in a variety of ways:
+
+Union
+
+The
union of
\(A\) and
\(B\) is the set
\(A \cup B = \{x \mid x \in A \text{ or } x \in B\}\text{.}\)
+
Intersection
+
+The
intersection of
\(A\) and
\(B\) is the set
\(A \cap B = \{x \mid x \in A \text{ and } x \in B\}\text{.}\)
+
Difference
+
+The
set difference \(A-B\) is the set
\(A - B = \{x \mid x \in A \text{ and } x \notin B\}\text{.}\)
+
Complement
+
+The
complement of
\(A\) is the set
\(A^c = \{x \in \Omega \mid x \notin A\}\text{.}\)
+
Empty Set
+
+The
empty set , usually written
\(\emptyset\) or
\(\{\}\text{,}\) is the set which contains no elements.
+
+
+
It’s useful sometimes to draw pictures representing sets...
+
+Definition 1.1.5 .
+
+Two sets
\(A\) and
\(B\) are
disjoint if
\(A \cap B = \emptyset\text{.}\)
+
+Definition 1.1.6 .
+
+Given a finite set
\(A\text{,}\) the
cardinality of
\(A\text{,}\) written
\(|A|\text{,}\) is the number of elements in
\(A\text{.}\)
+
+Exercises Exercises
+
+1. Consider the sets
\(A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\text{,}\) \(B = \{2, 4, 9, 10, 12, 14, 19\}\text{,}\) and
\(C = \{9, 10, 11, 14, 16, 17, 20\}\text{,}\) which are all subsets of
\(\Omega = \{1, 2, 3, \dotsc, 20\}\text{.}\)
+
(a) Find
\(A - (B \cap C)\text{.}\)
+
Answer .
+
+
\(\{1, 2, 3, 4, 5, 6, 7, 8\}\)
+
+
+
(b) Find
\(|A|\text{,}\) \(|B|\text{,}\) \(|C|\text{,}\) \(|A\cup B|\text{,}\) \(|A \cap B|\text{,}\) \(|B\cap C|\text{,}\) \(|A\cap C|\text{,}\) and
\(|A\cup B\cup C|\text{.}\) Is it true that the size of the union of sets is equal to the sum of the sizes of the individual sets?
+
Answer .
+
+
\(|A| = 10\text{,}\) \(|B| = 7\text{,}\) \(|C| = 7\text{,}\) \(|A \cup B| = 13\text{,}\) \(|A \cap B| = 4\text{,}\) \(|B\cap C| = 3\text{,}\) \(|A\cap C| = 2\text{,}\) \(|A\cup B\cup C| = 17\text{.}\) In particular, note that
\(|A\cup B| = 13 \neq 10 + 7 = |A| + |B|\text{,}\) so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets.
+
+
+
(c) Find
\(A^c\) and
\((A\cup B)^c\text{.}\)
+
Answer .
+
+
\(A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}\text{,}\) \((A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}\text{.}\)
+
+
+
2. Suppose we roll a 6-sided die two times. List the set of all possible results. [Note: the result (2, 4)---rolling a 2 and then a 4---is different from the result
\((4, 2)\) ---rolling a 4 and then a 2.]
+
3. Suppose we flip a coin two times. List the set of all possible results. What about flipping three times? Four times? If we flip the coin 10 times, how many possible results will there be?
+
4. If we roll a 6-sided die ten times, how many possible results will there be?
+
+
Two Sample Tests
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+\newcommand{\Z}{\mathbb Z}
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+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 5.2 Two Sample Tests
+
+
+
+
Variance
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+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+
+
+\DeclareMathOperator{\Bin}{Bin}
+\DeclareMathOperator{\Geom}{Geom}
+\DeclareMathOperator{\Poiss}{Poiss}
+\DeclareMathOperator{\Exp}{Exp}
+
+
+\DeclareMathOperator{\E}{E}
+\DeclareMathOperator{\Var}{Var}
+\newcommand{\lt}{<}
+\newcommand{\gt}{>}
+\newcommand{\amp}{&}
+\definecolor{fillinmathshade}{gray}{0.9}
+\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
+\)
+
+
+
+Section 3.2 Variance
+
+
+
+Exercises Exercises
+
+1. Consider a random variable
\(X\) with probability distribution below. Find
\(\Var(X)\text{.}\)
+
Table 3.2.1.
+
+
+1
+0.1
+
+
+2
+0.05
+
+
+3
+0.2
+
+
+4
+0.15
+
+
+5
+0.15
+
+
+6
+0.1
+
+
+7
+0.05
+
+
+8
+0.1
+
+
+9
+0.1
+
+
2. Let
\(X\) be a random variable taking the values 1, 2, 3, 4, 5. Write a distribution table for
\(X\text{,}\) then use your table to write a distribution for
\(X^2\text{.}\) Then, find
\(\Var(X)\text{.}\)
+
3. Suppose we flip a coin
\(n = 100\) times, and let
\(N\) count the number of heads. If the coin comes up heads on a flip with probability
\(p = 0.4\text{,}\) what is
\(\Var(N)\text{?}\) What if
\(n = 80\) and
\(p = 0.6\text{?}\) What if
\(n = 200\) and
\(p = 0.5\text{?}\)
+
4. If
\(\E(X) = 3\text{,}\) \(\Var(X) = 2\text{,}\) what is
\(\E(X^2)\text{?}\)
+
5. A continuous random variable
\(X\) taking values in
\([0, 1]\) has p.d.f.
\(f(x) = 2x\text{.}\) What is
\(\Var(X)\text{?}\)
+
6. A continuous random variable
\(X\) taking values in
\([-1, 1]\) has p.d.f.
\(f(x) = \frac{3x^2}{2}\text{.}\) What is
\(\Var(X)\text{?}\)
+
7. A continuous random variable
\(X\) taking values in
\([1, 4]\) has p.d.f.
\(f(x) = \frac{4}{3x^2}\text{.}\) What is
\(\Var(X)\text{?}\)
+
8. A continuous random variable
\(X\) taking values in
\([1, 4]\) has p.d.f.
\(f(x) = k(x - \sqrt{x})\) for some constant
\(k\text{.}\) In a previous problem, you found the value of
\(k\text{.}\) Now, find
\(\Var(X)\text{.}\)
+
9. A continuous random variable
\(X\) taking values in
\([1, 2]\) has p.d.f.
\(\displaystyle{f(x) = \frac{1}{2}\left(\frac{1}{x^2} + x\right)}\text{.}\) Find
\(\Var(X)\text{.}\)
+
+