4.1.0 Continuous Random Variables and their Distributions
We have in fact already seen examples of continuous random variables before, e.g., Example 1.14. Let us look at the same example with just a little bit different wording.
Example
I choose a real number uniformly at random in the interval $[a,b]$, and call it $X$. By uniformly at random, we mean all intervals in $[a,b]$ that have the same length must have the same probability. Find the CDF of $X$.
- Solution
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As we mentioned, this is almost exactly the same problem as Example 1.14, with the difference
being, in that problem, we considered the interval from $1$ to $2$. In that example,
we saw that all individual points have probability $0$, i.e., $P(X=x)=0$ for all $x$. Also,
the uniformity implies that the probability of an interval of length $l$ in $[a,b]$ must be
proportional to its length:
$$P(X \in [x_1,x_2]) \propto (x_2-x_1), \hspace{20pt} \textrm{where }a \leq x_1 \leq x_2 \leq b.$$
Since $ P(X \in [a,b])=1$, we conclude
$$P(X \in [x_1,x_2]) =\frac{x_2-x_1}{b-a}, \hspace{20pt} \textrm{where }a \leq x_1 \leq x_2 \leq b.$$
Now, let us find the CDF. By definition $F_X(x)=P(X \leq x)$, thus we immediately have
$$F_X(x) =0, \hspace{20pt} \textrm{for } x < a,$$
$$F_X(x) =1, \hspace{20pt} \textrm{for } x \geq b.$$
For $a \leq x \leq b$, we have
$F_X(x)$ $=P(X \leq x)$ $=P(X \in [a,x])$ $=\frac{x-a}{b-a}.$
Thus, to summarize \begin{equation} \hspace{70pt} F_X(x) = \left\{ \begin{array}{l l} 0 & \quad \textrm{for } x < a \\ \frac{x-a}{b-a} & \quad \textrm{for }a \leq x \leq b\\ 1 & \quad \textrm{for } x > b \end{array} \right. \hspace{70pt} (4.1) \end{equation} Note that here it does not matter if we use "$ < $" or "$ \leq $", as each individual point has probability zero, so for example $P(X < 2)=P(X \leq 2)$. Figure 4.1 shows the CDF of $X$. As we expect the CDF starts at zero and ends at $1$.
Fig.4.1 - CDF for a continuous random variable uniformly distributed over $[a,b]$.
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As we mentioned, this is almost exactly the same problem as Example 1.14, with the difference
being, in that problem, we considered the interval from $1$ to $2$. In that example,
we saw that all individual points have probability $0$, i.e., $P(X=x)=0$ for all $x$. Also,
the uniformity implies that the probability of an interval of length $l$ in $[a,b]$ must be
proportional to its length:
$$P(X \in [x_1,x_2]) \propto (x_2-x_1), \hspace{20pt} \textrm{where }a \leq x_1 \leq x_2 \leq b.$$
Since $ P(X \in [a,b])=1$, we conclude
$$P(X \in [x_1,x_2]) =\frac{x_2-x_1}{b-a}, \hspace{20pt} \textrm{where }a \leq x_1 \leq x_2 \leq b.$$
Now, let us find the CDF. By definition $F_X(x)=P(X \leq x)$, thus we immediately have
$$F_X(x) =0, \hspace{20pt} \textrm{for } x < a,$$
$$F_X(x) =1, \hspace{20pt} \textrm{for } x \geq b.$$
For $a \leq x \leq b$, we have
One big difference that we notice here as opposed to discrete random variables is that the CDF is a continuous function, i.e., it does not have any jumps. Remember that jumps in the CDF correspond to points $x$ for which $P(X=x) > 0$. Thus, the fact that the CDF does not have jumps is consistent with the fact that $P(X=x)=0$ for all $x$. Indeed, we have the following definition for continuous random variables.
A random variable $X$ with CDF $F_X(x)$ is said to be continuous if $F_X(x)$ is a continuous function for all $x \in \mathbb{R}$.
We will also assume that the CDF of a continuous random variable is differentiable almost everywhere in $\mathbb{R}$.
