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Suppose a switch is pulled to restart power in a house every week between 0 and 4 times. Here is the cumulative distribution function for the number of times the switch is pulled:

           0,    if x < 0

           0.17, if 0 <= x < 1

 F(x) = {  0.53, if 1 <= x < 2

           0.84, if 2 <= x < 3

           0.97, if 3 <= x < 4

           1,    if x >= 4

How do I find the expected number of times that the switch needs to be pulled during the week? I know that I can find the expected value of the cumulative distribution function, but how can I use that (or can I even use it?) to find the expected number of times the switch is pulled?

Here is the expected value:

E[X] = (0 * 0.17) + (1 * 0.36) + (2 * 0.31) + (3 * 0.13) + (4 * 0.03) = 1.49.

1 Answers 1

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There is nothing wrong with what you have done.   It is a way to evaluate the expectation from the cumulative distribution function.   Correctly done at that.

However, you may be looking for this:

If $X$ is a strictly non-negative integer valued discrete random variable with cumulative distribution function $F_X(\;)$, then the following may be used:$$\mathsf E(X)= \sum_{x=0}^\infty (1-F_X(x))$$

in this case $\mathsf E(X)=(1-0.17)+(1-0.53)+(1-0.84)+(1-0.97)+(1-1)+\ldots \\\qquad = 1.49$

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    I am a little confused: it was my understanding that the expected value was not the same as the expected number of times the switch would be pulled. Am I incorrect? Can is the expected number of times the switch is going to be pulled 1? (Because 1 <= 1.49 < 2?) – 2017-02-09
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    The random variable is the count of times the switch is pulled. The expected value of this random variable is thus, the *expected count of times the switch would be pulled*. Both calculations agree that this is $1.49$. (The expected value *need not* be an integer value even if the random variable is strictly integer valued.) – 2017-02-09
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    Awesome, thanks for your help! – 2017-02-09