Expected Value of a Continuous Function

  • Understand conditions for continuous probability distribution function.

  • Calculate the expected value of a continuous function.

\[\mu = E(X) = \int_{-\infty}^\infty x_i P(x_i)\]

Solved Examples

Solved Example:

9-3-01

Let the random variable X denotes the time a person waits for a bus to arrive. X varies according to the graph shown below. What is the expected value of waiting, in minutes, for a person on this bus stop?

Solution:

9.3-01

Method I: Using fundamental definition of expected value. Let us find the equation of declining line.

$\mathrm{Slope} = m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{0-0.5}{3 - 1} = \dfrac{-1}{4}$ and the line passes through ($x_1$, $y_1$) = (3,0).

Using, $y - y_1 = m (x- x_1) \mathrm{\ We\ get\ } y = -0.25x + 0.75$

\[P(x) = \begin{cases} \text{0.5,} &\quad 0 \leq x \leq 1\\ \text{-0.25x + 0.75} &\quad 1 \leq x \leq 3 \\ \end{cases} \] \begin{align*} E(X) &= \int_0^3 x P(x) dx\\ &= \int_0^1 0.5x dx + \int_1^3 x (-0.25x + 0.75) dx\\ &= 0.5 \left[\dfrac{x^2}{2}\right]_0^1 + (-0.25)\left[\dfrac{x^3}{3}\right]_1^3 + 0.75\left[\dfrac{x^2}{2}\right]_1^3\\ &= 0.25 - (0.25)(8.67) + 0.75(4)\\ &= 0.25 - 2.167 + 3 = 1.083 \end{align*}

Method II: Using concept of Centroid.

We will calculate the x coordinate of the centroid of the combined shape.

\begin{align*} \mathrm{Rectangle}: \\A_1 &= 0.5,\\ x_1 &= 0.5\\ \mathrm{Triangle}: \\A_2&= 0.5 \times 2 \times 0.5 = 0.5,\\ x_2 &= 1 + \dfrac{1}{3} (2) = 1.67 \end{align*} \begin{align*} \bar{x} &= \dfrac{A_1 x_1 + A_2 x_2}{A_1 + A_2}\\ &= \dfrac{(0.5)(0.5)+(0.5)(1.667)}{(0.5)+(0.5)}\\ &= 1.083 \end{align*}

Correct Answer: B