4.4: Expected Value and Variance of Continuous Random Variables

Definition $$\PageIndex{1}$$

If $$X$$ is a continuous random variable with density function $$f(x)$$, then the expected value (or mean) of $$Y$$ is given by

$$\mu = \mu_X = E[X] = \int\limits^{\infty}_{-\infty}\! x\cdot f(x)\, dx.\notag$$

The formula for the expected value of a continuous random variable is the continuous analogue of the expected value of a discrete random variable, where instead of summing over all possible values we integrate, recall section 3.7. This interpretation of the expected value as a weighted average explains why it is also referred to as the mean of the random variable.

Example $$\PageIndex{1}$$

Consider again the context of Example 17, where we defined the continuous random variable $$X$$ to denote the time a person waits for an elevator to arrive. The pdf of $$X$$ was given by
$$f(x) = \left\{\begin{array}{l l} x, & \text{for}\ 0\leq x\leq 1 \\ 2-x, & \text{for}\ 1< x\leq 2 \\ 0, & \text{otherwise} \end{array}\right.\notag$$
Applying Definition 4.4.1, we compute the expected value of $$X$$:
$$E[X] = \int\limits^1_0\! x\cdot x\, dx + \int\limits^2_1\! x\cdot (2-x)\, dx = \int\limits^1_0\! x^2\, dx + \int\limits^2_1\! (2x - x^2)\, dx = \frac{1}{3} + \frac{2}{3} = 1.\notag$$
Thus, we expect a person will wait 1 minute for the elevator on average. Figure 1 demonstrates the graphical representation of the expected value as the center of mass of the pdf. Figure 1: Graph of $$f$$: The red arrow represents the center of mass, or the expected value of $$X$$.

If continuous random variable $$X$$ has a normal distribution with parameters $$\mu$$ and $$\sigma$$, then $$E[X] = \mu$$. The normal case is why the notation $$\mu$$ is often used for the expected value. Again, this fact can be derived using Definition 4.4.1; however, the integral calculation requires many tricks.

The expected value may not be exactly equal to a parameter of the probability distribution, but rather it may be a function of the parameters as the next example with the uniform distribution shows.

Example $$\PageIndex{2}$$

Suppose the random variable $$X$$ has a uniform distribution on the interval $$[a,b]$$. Then the pdf of $$X$$ is given by
$$f(x) = \frac{1}{b-a}, \quad\text{for}\ a\leq x\leq b.\notag$$
Applying Definition 4.4.1, we compute the expected value of $$X$$:
$$E[X] = \int\limits^b_a\! x\cdot\frac{1}{b-a}\, dx = \frac{b^2 - a^2}{2}\cdot\frac{1}{b-a} = \frac{(b-a)(b+a)}{2}\cdot\frac{1}{b-a} = \frac{b+ a}{2}.\notag$$
Thus, the expected value of the uniform$$[a,b]$$ distribution is given by the average of the parameters $$a$$ and $$b$$, or the midpoint of the interval $$[a,b]$$. This is readily apparent when looking at a graph of the pdf. Since the pdf is constant over $$[a,b]$$, the center of mass is simply given by the midpoint.

If $$X$$ is a continuous random variable with pdf $$f(x)$$, then the expected value of $$Y$$ is given by CONTINUOUS CASE

$$E[Y] = \int\limits^{\infty}_{-\infty}\! g(x)\cdot f(x)\, dx.\notag$$

2. If $$X_1, \ldots, X_n$$ are continuous random variables with joint density function $$p(x_1, \ldots, x_n)$$, then the expected value of $$Y$$ is given by CONTINUOUS CASE
$$E[Y] = \int\limits^{\infty}_{-\infty}\!\cdots\int\limits^{\infty}_{-\infty}\! g(x_1, \ldots, x_n)\cdot f(x_1, \ldots, x_n)\, dx_1\, \ldots\, dx_n.\notag$$