Q. 7.6 - A First Course in Probability

Question

In the text, we noted that

$E\left[\sum _{i=1}^{\infty }{X}_i\right]=\sum _{i=1}^{\infty }E\left[X_i\right]$

when the $Xi$ are all nonnegative random variables. Since

an integral is a limit of sums, one might expect that $E\left[{\int }_0^{\infty }X\left(t\right)dt\right]={\int }_0^{\infty }E\left[X\right(t\left)\right]dt$

whenever $X\left(t\right),0\le t<\infty ,$ are all nonnegative random

variables; this result is indeed true. Use it to give another proof of the result that for a nonnegative random variable $X$ ,

$E\left[X\right)={\int }_0^{\infty }P(X>t\}dt$

Hint: Define, for each nonnegative $t$, the random variable

$X\left(t\right)$ by

$X\left(t\right)= 1 if t<X \backslash \backslash 0 if t \ge X$

Now relate $4q$

${\int }_0^{\infty }X\left(t\right)dt\text{ to }X$

Step-by-Step Solution

Answer

The result that for a nonnegative random variable $X$ is $E\left(X\right)={\int }_0^{\infty }P(X>t)dt$ claimed as two equal expressions.

Step 1: Given Information

 All nonnegative random variables is an integral is a limit of sums as $E\left[{\int }_0^{\infty }X\left(t\right)dt\right]={\int }_0^{\infty }E\left[X\right(t\left)\right]dt$.

Step 2: Explanation

Define random variables as given in the hint

Observe that $X\left(t\right)$ is in fact the characteristic function that $X$ falls in $(t,\infty )$. Now, on the left side we have that,

$E\left({\int }_0^{\infty }X\left(t\right)dt\right)=E\left({\int }_0^{\infty }{\chi }_{(t,\infty )}\left(X\right)dt\right)=E\left(X\right)$

Step 3: Explanation

On the right side we have that,

Since these two beginning expressions are equal, it has to be

$E\left(X\right)={\int }_0^{\infty }P(X>t)dt$

which has been claimed.

Step 4: Final answer

The result that for a nonnegative random variable $X$ is $E\left(X\right)={\int }_0^{\infty }P(X>t)dt$ claimed as two equal expressions.