For continuous random variables, expected values are defined in a similar way, but with the double summation replaced by a double integral.

Definition

If \(X\) and \(Y\) are continuous random variables with joint probability density function \(f(x,y)\), then the expected value of a function of the variables, \(g(X,Y)\), is defined to be

\[ E[g(X,Y)] \;=\; \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty} {g(x,y) \times f(x,y)} \;dx} \;dy \]

This again corresponds to each possible value of \(g(x,y)\) being "weighted" by its probability density, with the most likely values of the variables contributing most to the expected value.

Example

A point is randomly selected within a unit square. What is the expected area of the rectangle with it and the origin as corners?

What is its variance?

 


\(X\) and \(Y\) are independent \(\UniformDistn(0,1)\) random variables, so their joint probability density function is

\[ f(x,y) \;=\; 1 \qquad \text{for } 0 \le x \le 1 \text{ and } 0 \le y \le 1 \]

The expected area is

\[ \begin{align} E[XY] \;&=\; \int_0^1 {\int_0^1 xy\;dx}\;dy \\ &=\; \int_0^1 x \;dx \; \times\; \int_0^1 y \;dy \\ &=\; \left[\frac{x^2}{2}\right]_0^1 \; \times\; \left[\frac{y^2}{2}\right]_0^1 \\ &=\; \left(\frac 1 2\right)^2 \;=\; \frac 1 4 \end{align} \]

To find the variance of \(XY\), we first find \(E\left[(xy)^2\right]\).

\[ \begin{align} E\left[(XY)^2\right] \;&=\; \int_0^1 {\int_0^1 x^2y^2\;dx}\;dy \\ &=\; \int_0^1 x^2 \;dx \; \times\; \int_0^1 y^2 \;dy \\ &=\; \left[\frac{x^3}{3}\right]_0^1 \; \times\; \left[\frac{y^3}{3}\right]_0^1 \\ &=\; \left(\frac 1 3\right)^2 \;=\; \frac 1 9 \end{align} \]

The variance is found from

\[ \Var(XY) \;=\; E\left[(XY)^2\right] - \left(E[XY]\right)^2 \;=\; \frac 1 9 - \frac 1{16} \;=\; \frac 7{144} \]