4.1.2 Expected Value and Variance
As we mentioned earlier, the theory of continuous random variables is very similar to the theory of discrete random variables. In particular, usually summations are replaced by integrals and PMFs are replaced by PDFs. The proofs and ideas are very analogous to the discrete case, so sometimes we state the results without mathematical derivations for the purpose of brevity.
Remember that the expected value of a discrete random variable can be obtained as $$EX=\sum_{x_k \in R_X} x_k P_X(x_k).$$ Now, by replacing the sum by an integral and PMF by PDF, we can write the definition of expected value of a continuous random variable as
Example
Let $X \sim Uniform(a,b)$. Find $EX$.
- Solution
-
As we saw, the PDF of $X$ is given by
\begin{equation}
\nonumber f_X(x) = \left\{
\begin{array}{l l}
\frac{1}{b-a} & \quad a < x < b\\
0 & \quad x < a \textrm{ or } x > b
\end{array} \right.
\end{equation}
so to find its expected value, we can write
$EX$ $= \int_{-\infty}^{\infty} xf_X(x)dx$ $=\int_{a}^{b} x (\frac{1}{b-a}) dx$ $=\frac{1}{b-a} \bigg[ \frac{1}{2}x^2 \bigg]_{a}^{b} $ $=\frac{a+b}{2}.$
This result is intuitively reasonable: since $X$ is uniformly distributed over the interval $[a,b]$, we expect its mean to be the middle point, i.e., $EX=\frac{a+b}{2}$.
-
As we saw, the PDF of $X$ is given by
\begin{equation}
\nonumber f_X(x) = \left\{
\begin{array}{l l}
\frac{1}{b-a} & \quad a < x < b\\
0 & \quad x < a \textrm{ or } x > b
\end{array} \right.
\end{equation}
so to find its expected value, we can write
Example
Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} 2x & \quad 0 \leq x \leq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Find the expected value of $X$.
- Solution
-
We have
$EX$ $= \int_{-\infty}^{\infty} xf_X(x)dx$ $=\int_{0}^{1} x (2x) dx$ $=\int_{0}^{1} 2x^2 dx$ $=\frac{2}{3}.$
-
We have
Expected Value of a Function of a Continuous Random Variable
Remember the law of the unconscious statistician (LOTUS) for discrete random variables: $$\hspace{70pt} E[g(X)]=\sum_{x_k \in R_X} g(x_k)P_X(x_k) \hspace{70pt} (4.2)$$ Now, by changing the sum to integral and changing the PMF to PDF we will obtain the similar formula for continuous random variables.
As we have seen before, expectation is a linear operation, thus we always have
- $E[aX+b]=aEX+b$, for all $a,b \in \mathbb{R}$, and
- $E[X_1+X_2+...+X_n]=EX_1+EX_2+...+EX_n$, for any set of random variables $X_1, X_2,...,X_n$.
Example
Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} x+\frac{1}{2} & \quad 0 \leq x \leq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Find $E(X^n)$, where $n \in \mathbb{N}$.
- Solution
-
Using LOTUS we have
$E[X^n]$ $=\int_{-\infty}^{\infty} x^n f_X(x) dx$ $= \int_{0}^{1} x^n (x+\frac{1}{2}) dx$ $= \bigg[\frac{1}{n+2}x^{n+2}+\frac{1}{2(n+1)}x^{n+1} \bigg]_{0}^{1}$ $=\frac{3n+4}{2(n+1)(n+2)}.$
-
Using LOTUS we have
Variance
Remember that the variance of any random variable is defined as $$\textrm{Var}(X)=E\big[(X-\mu_X)^2\big]=EX^2-(EX)^2.$$ So for a continuous random variable, we can write
| $$\textrm{Var}(X)$$ | $$=E\big[(X-\mu_X)^2\big]=\int_{-\infty}^{\infty} (x-\mu_X)^2 f_X(x)dx$$ |
| $$=EX^2-(EX)^2=\int_{-\infty}^{\infty} x^2 f_X(x)dx-\mu_X^2$$ |
Also remember that for $a,b \in \mathbb{R}$, we always have $$\textrm{Var}(aX+b)=a^2 \textrm{Var}(X).\hspace{70pt} (4.4)$$
Example
Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} \frac{3}{x^4} & \quad x \geq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Find the mean and variance of $X$.
- Solution
-
$E[X]$ $=\int_{-\infty}^{\infty} x f_X(x) dx$ $= \int_{1}^{\infty} \frac{3}{x^3} dx$ $= \bigg[-\frac{3}{2}x^{-2} \bigg]_{1}^{\infty}$ $=\frac{3}{2}.$
Next, we find $EX^2$ using LOTUS,$E[X^2]$ $=\int_{-\infty}^{\infty} x^2 f_X(x) dx$ $= \int_{1}^{\infty} \frac{3}{x^2} dx$ $= \bigg[-3x^{-1} \bigg]_{1}^{\infty}$ $=3.$
Thus, we have $$\textrm{Var}(X)=EX^2-(EX)^2=3-\frac{9}{4}=\frac{3}{4}.$$
-
