STAT 400

Wed. February 26th, 2020

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Continuous Random Variables

We call a random variable $X$ a continuous random variable if it satisfies these two properties:

• $X$ can be any value on an entire interval (like $[0,1]$ or $(-\infin,\infin)$)
• For any $a$, $P(X=a)=0.$ (This is like taking an integral of a function over an empty interval – the result will always be 0 regardless of the function.)

Instead of studying cases of $P(X=0)$, we instead focus on $P(a\le X\le b)$. Using "$\le$" and "$<$" is equivalent for continuous random variables, unlike the discrete case.

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Ex. Consider a chance wheel. Let $X$ be the angle that the arrow is pointing to after the wheel is spun. In this case, $X$ can take any value on the interval $[0,2\pi)$, so it’s continuous.

You can see how it’s pointless to ask what the probability is that $X$ lands exactly on a specific angle, since it’ll always be 0. Instead, we can calculate expressions like: $\begin{aligned} P(0\le X\le \pi)&=\frac{1}{2}=\frac{\pi}{2\pi}\\ P(a\pi \le X\le b\pi)&=\frac{(b-a)\pi}{2\pi}=\frac{b-a}{2}\\ P(a\pi \le X)=P(a\pi \le X \le 2\pi)&=\frac{2-a}{2} \end{aligned}$

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Probability Density Function

We can add together the probabilities of $X$ being in consecutive intervals to get the probability of $X$ being in the union of the two intervals: $P(a_1\le X\le a_2)+P(a_2\le X\le a_3)=P(a_1\le X\le a_3).$ Integrals also have this property as well, so it could be useful to find a way to represent our probabilities using them.

We’re looking for a function $f$ such that, for any $a\le b$, $P(a\le X\le b)=\int^{b}_{a}f(x)dx.$ We call $f$ the probability density function (pdf), probability distribution, or density curve of $X$. This turns out to be the analogue of the probability mass function for continuous random variables.

Most of the continuous distributions we’ll study don’t really have easy-to-explain motivations behind them, but we can still start with a pdf function and derive all other properties from there (expected value, variance, cdf).

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Ex. Going back to our chance wheel example, $P(a_1\le X\le a_2)=\frac{a_2-a_1}{2\pi}$ for $0< a_1< a_2< 2\pi$.
In this case, the probability density function $f(x)$ for $X$ should be $f(x)=\begin{cases} \frac{1}{2\pi}&x\in[0,2\pi)\\ 0&\text{otherwise} \end{cases}$

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PDF Remarks

1. $f(x)\ge 0$ for all $x$.
2. $\int_{-\infin}^{\infin}f(x)dx=1.$
3. $P(X=a)\ne f(a)$, don’t mess this up!
4. $P(a-\frac{1}{2}\Delta \le X\le a+\frac{1}{2}\Delta)\approx \Delta f(a)$

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Arbitrary PDFs

Ex. Suppose the pdf of a random variable $X$ is $f(x)=\begin{cases} kx^2&0\le x\le 2\\ 0&\text{otherwise} \end{cases}$

1. What is $k$? We know that the integral over all values of $f(x)$ has to equal $1$. $\begin{aligned} \int_{-\infin}^{\infin}f(x)dx&=\int_0^2 kx^2dx\\ &=\frac{1}{3}kx^3\big|_0^2\\&=\frac{8}{3}k=1\\&\rightarrow k=\frac{3}{8}. \end{aligned}$
2. Calculate $P(X\le 1)$. $\begin{aligned} P(X\le 1)&=\int_{-\infin}^1 f(x)dx\\ &=\int_0^1 \frac{3}{8}x^2dx\\& =\frac{1}{8}x^3\big|_0^1=\frac{1}{8}.\\ \end{aligned}$

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Uniform Distribution

Definition. A continuous random variable $X$ is said to be a uniform distribution on an interval $[A,B]$ if its probability density function is $f(x;A,B)=\begin{cases} \frac{1}{B-A}&x\in[A,B]\\ 0&\text{otherwise} \end{cases}$ Here, we say $X$ follows $\text{unif}(A,B)$.

(Note that our chance wheel example from before follows this distribution.)