3.2.3 Functions of Random Variables

If $X$ is a random variable and $Y=g(X)$, then $Y$ itself is a random variable. Thus, we can talk about its PMF, CDF, and expected value. First, note that the range of $Y$ can be written as $$R_Y=\{g(x) | x \in R_X \}.$$ If we already know the PMF of $X$, to find the PMF of $Y=g(X)$, we can write \begin{align}%\label{} \nonumber P_Y (y) &= P(Y=y)\\ \nonumber &= P(g(X)=y)\\ \nonumber &= \sum_{x:g(x)=y} P_X(x)\\ \end{align}
Let's look at an example.



Example

Let $X$ be a discrete random variable with $P_X(k)=\frac{1}{5}$ for $k=-1,0,1,2,3$. Let $Y=2|X|$. Find the range and PMF of $Y$.

  • Solution
    • First, note that the range of $Y$ is
      $R_Y$ $=\{2|x| \textrm{ where } x \in R_X \}$
      $=\{0,2,4,6\}$.

      To find $P_Y(y)$, we need to find $P(Y=y)$ for $y=0,2,4,6$. We have

      $P_Y(0)$ $=P(Y=0)=P(2|X|=0)$
      $=P(X=0)= \frac{1}{5}$;
      $P_Y(2)$ $=P(Y=2)=P(2|X|=2)$
      $=P\big((X=-1) \textrm{ or } (X=1) \big)$
      $=P_X(-1)+ P_X(1)= \frac{1}{5}+\frac{1}{5}=\frac{2}{5}$;
      $P_Y(4)$ $=P(Y=4)=P(2|X|=4)$
      $=P(X=2)+P(X=-2)= \frac{1}{5}$;
      $P_Y(6)$ $=P(Y=6)=P(2|X|=6)$
      $=P(X=3)+P(X=-3)= \frac{1}{5}$.

      So, to summarize, \begin{equation} \nonumber P_Y(k) = \left\{ \begin{array}{l l} \frac{1}{5} & \quad \text{for } k=0,4,6\\ \frac{2}{5} & \quad \text{for } k=2\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}


Expected Value of a Function of a Random Variable (LOTUS)

Let $X$ be a discrete random variable with PMF $P_X(x)$, and let $Y=g(X)$. Suppose that we are interested in finding $EY$. One way to find $EY$ is to first find the PMF of $Y$ and then use the expectation formula $EY=E[g(X)]=\sum_{y \in R_Y} y P_Y(y)$. But there is another way which is usually easier. It is called the law of the unconscious statistician (LOTUS).

Law of the unconscious statistician (LOTUS) for discrete random variables:
$$\hspace{60pt} E[g(X)]=\sum_{x_k \in R_X} g(x_k)P_X(x_k) \hspace{80pt} (3.2)$$

You can prove this by writing $EY=E[g(X)]=\sum_{y \in R_Y} y P_Y(y)$ in terms of $P_X(x)$. In practice it is usually easier to use LOTUS than direct definition when we need $E[g(X)]$.



Example

Let $X$ be a discrete random variable with range $R_X=\{0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi \}$, such that $P_X(0)=P_X(\frac{\pi}{4})=P_X(\frac{\pi}{2})=P_X(\frac{3\pi}{4})=P_X(\pi)=\frac{1}{5}$. Find $E[\sin (X)]$.

  • Solution
    • Using LOTUS, we have

      $E[g(X)]$ $=\sum_{x_k \in R_X} g(x_k)P_X(x_k)$
      $=\sin(0) \cdot \frac{1}{5}+\sin(\frac{\pi}{4}) \cdot \frac{1}{5}+\sin(\frac{\pi}{2}) \cdot \frac{1}{5}+ \sin(\frac{3\pi}{4}) \cdot \frac{1}{5}+ \sin(\pi) \cdot \frac{1}{5}$
      $=0 \cdot \frac{1}{5}+\frac{\sqrt{2}}{2} \cdot \frac{1}{5}+1 \cdot \frac{1}{5}+ \frac{\sqrt{2}}{2} \cdot \frac{1}{5}+ 0 \cdot \frac{1}{5}$
      $=\frac{\sqrt{2}+1}{5}$.



Example

Prove $E[aX+b]=aEX+b$ (linearity of expectation).

  • Solution
    • Here $g(X)=aX+b$, so using LOTUS we have

      $E[aX+b]$ $=\sum_{x_k \in R_X} (ax_k+b)P_X(x_k)$
      $=\sum_{x_k \in R_X} ax_kP_X(x_k)+ \sum_{x_k \in R_X} bP_X(x_k)$
      $=a \sum_{x_k \in R_X} x_kP_X(x_k)+ b\sum_{x_k \in R_X} P_X(x_k)$
      $=a EX+ b$.






The print version of the book is available through Amazon here.

Book Cover