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.
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
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.
Let \(X\) take values in the interval \([a, b]\text{.}\) We say that \(X\) has the uniform distribution if, for any subinterval \([c, d] \subset [a, b]\text{,}\) the probability that \(X\) lies within \([c, d]\) is equal to the proportion of the total sample space taken by \([c, d]\text{.}\) That is:
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 \(k\text{:}\)
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.
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*}
(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 \(t\text{.}\))
This equation may look familiar from Calculus 1. If you review the First Fundamental Theorem of Calculus, you’ll see precisely the relationship between the cdf \(F(x)\) and the pdf \(f(x)\text{:}\)
\begin{gather*}
F'(x) = f(x).
\end{gather*}
Let’s see how we can put this relationship to use.
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
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.
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.
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.
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. We’ll write \(X \sim \Norm(\mu, \sigma^2)\text{.}\)
In the specific case that \(\mu = 0\) and \(\sigma^2 = 1\text{,}\) we call the resulting distribution the standard normal distribution. We’ll use the notation \(\phi(x)\) for the standard normal density function, and \(\Phi(x)\) for the standard normal cumulative distribution function.
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).
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 \(\Pr\left(1 \leq X \leq \frac{3}{2}\right)\text{.}\)
