Skip to main content
Contents Embed Dark Mode Prev Up Next \(\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb Z}
\newcommand{\Q}{\mathbb Q}
\newcommand{\R}{\mathbb R}
\DeclareMathOperator{\Bin}{Bin}
\DeclareMathOperator{\Geom}{Geom}
\DeclareMathOperator{\Poiss}{Poiss}
\DeclareMathOperator{\Exp}{Exp}
\DeclareMathOperator{\E}{E}
\DeclareMathOperator{\Var}{Var}
\DeclareMathOperator{\Cov}{Cov}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section Tuesday, Jan 27
This is an outline of the topics we covered in class. These notes are
not a substitute for your own note-taking. I highly recommend that you take your own notes during class. If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
Subsection Continuous Distributions
Example 55 .
Let \(X\in [0, 1]\) with \(f(x) = kx^{3/2}\text{.}\) Find \(k\text{.}\)
\begin{align*}
1 = \int_0^1 f(x)\ dx \amp = \int_0^1 kx^{3/2} dx = \frac{2kx^{5/2}}{5}\bigg|_0^1 = \frac{2k}{5} - 0 \\
\Rightarrow \quad 1 \amp = \frac{2k}{5} \quad \Rightarrow \quad k = \frac{5}{2}
\end{align*}
Then, we can calculate probabilities, e.g.:
\begin{gather*}
\Pr\left(X \lt \frac{1}{2}\right) = \int_0^{1/2} \frac{5}{2}x^{3/2}\ dx = \frac{5}{2}\cdot \frac{2x^{5/2}}{5}\bigg|_0^{1/2} = \left(\frac{1}{2}\right)^{5/2} \approx 0.177.
\end{gather*}
Note: a pdf outputs probability densities, not probabilities. To get probabilities, we must integrate.
Definition 56 .
Let \(X\) be a continuous random variable with values in \([a, b]\text{.}\) The cumulative distribution function (cdf) is:
\begin{gather*}
F(x) = \Pr(X \leq x).
\end{gather*}
To calculate it:
\begin{gather*}
F(x) = \int_a^x f(t)\ dt.
\end{gather*}
Example 57 .
Let \(X \in [0, 1], f(x) = \frac{5}{2}x^{3/2}\text{.}\) Then:
\begin{align*}
F(x) = \int_0^x f(t)\ dt \amp = \int_0^x \frac{5}{2}t^{3/2}\ dt = \frac{5}{2}\cdot \frac{2t^{5/2}}{5}\bigg|_0^x \\
F(x) \amp = x^{5/2}
\end{align*}
Then, we can calculate probabilities, e.g.:
\begin{align*}
\Pr\left(X \lt \frac{1}{2}\right) \amp = F\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{5/2} \approx 0.177 \\
\Pr(0.2 \leq X \leq 0.6) \amp = F(0.6) - F(0.2) = (0.6)^{5/2} - (0.2)^{5/2} \approx 0.261
\end{align*}
Given \(f(x)\text{,}\) we can find \(F(x)\) by calculating:
\begin{gather*}
F(x) = \int_a^x f(t)\ dt.
\end{gather*}
Given \(F(x)\text{,}\) we can find \(f(x)\) by calculating:
\begin{gather*}
f(x) = F'(x).
\end{gather*}
Example 58 .
Suppose a machine needs repairs on average twice per month. Let
\(T\) be the time until repair. This is a Poisson process (
\(\lambda = 2\) per month).
Definition 59 .
\(T\) is said to have the exponential distribution with parameter \(\lambda\text{.}\) We’ll write \(T \sim \Exp(\lambda)\text{.}\) The exponential density function is:
\begin{gather*}
f(t) = \lambda e^{-\lambda t}, \quad t \geq 0.
\end{gather*}
Example 60 .
Let \(\lambda = 2, f(t) = 2e^{-2t}, t\geq 0\text{.}\) Then:
\begin{align*}
F(t) \amp = \int_0^t f(x)\ dx = \int_0^t 2e^{-2x}\ dx = -e^{-2x}\bigg|_0^x \\
\amp = -e^{-2t} - (-e^0) = 1 - e^{-2t}.
\end{align*}
So, e.g.:
\begin{align*}
\Pr\left(T \leq \frac{3}{4}\right) \amp = F\left(\frac{3}{4}\right) = 1 - e^{-3/2} \approx 0.777 \\
\Pr(T \gt 1) \amp = 1 - \Pr(T \leq 1) = 1 - F(1) \\
\amp = 1 - (1 - e^{-2}) = e^{-2} \approx 0.135.
\end{align*}
Subsection Joint Distributions
Definition 63 .
Let
\(X\) take values
\(x_1, x_2, \dotsc, x_n\) and
\(Y\) take values
\(y_1, y_2, \dotsc, y_m\text{.}\) The
joint distribution of
\(X\) and
\(Y\) is the collection of all values
\(\Pr(X = x_i, Y = y_j)\) for every
\(i, j\) combination. The separarte distributions for
\(X\) and
\(Y\) are called
marginal distributions .
Example 64 .
Suppose
\(X \in \{1, 2, 3\}, Y \in \{0, 1\}\text{,}\) with joint distribution below.
Table 65. Example Joint Distribution
\(X = 1\) \(X = 2\) \(X = 3\)
\(Y = 0\) 0.1
0.15
0.05
\(Y = 1\) 0.2
0.2
0.3
We find the marginal distribution for \(X\) by summing along the columns:
\begin{align*}
\Pr(X = 1) \amp = 0.3 \\
\Pr(X = 2) \amp = 0.35 \\
\Pr(X = 3) \amp = 0.35
\end{align*}
We find the marginal distribution for \(Y\) by summing along the rows:
\begin{align*}
\Pr(Y = 0) \amp = 0.3 \\
\Pr(Y = 1) \amp = 0.7
\end{align*}
Definition 66 .
Random variables \(X, Y\) are independent if:
\begin{gather*}
\Pr(X = x_i, Y = y_j) = \Pr(X = x_i)\Pr(Y = y_j)
\end{gather*}
for every \(i, j\) combination.
Example 67 .
In the previous example:
\begin{gather*}
\Pr(X = 2, Y = 1) = 0.2 \neq (0.35)(0.7) = \Pr(X = 2)\Pr(Y = 1),
\end{gather*}
so \(X, Y\) are not independent.
Example 68 .
Let
\(X, Y\) indicate heads on the first and second flip, respectively, of a fair coin. Then:
Table 69. Joint Distribution for Indicator Random Variables
\(X = 0\) \(X = 1\)
\(Y = 0\) 1/4
1/4
\(Y = 1\) 1/4
1/4
Then the marginal distributions are:
\begin{align*}
\Pr(X = 0) \amp = \frac{1}{2} \amp \Pr(Y = 0) \amp = \frac{1}{2} \\
\Pr(X = 1) \amp = \frac{1}{2} \amp \Pr(Y = 1) \amp = \frac{1}{2}
\end{align*}
Then, for any \(a, b\text{:}\)
\begin{gather*}
\Pr(X = a, Y = b) = \frac{1}{4} = \frac{1}{2}\cdot \frac{1}{2} = \Pr(X = a)\Pr(Y = b).
\end{gather*}
So \(X, Y\) are independent.
