Question

Let \(X\) be a random variable of the discrete type with pmf \(p(x)\) that is positive on the nonnegative integers and is equal to zero elsewhere. Show that $$ E(X)=\sum_{x=0}^{\infty}[1-F(x)] $$ where \(F(x)\) is the cdf of \(X\).

Short answer

The equality \( E(X)=\sum_{x=0}^{\infty}[1-F(x)] \) is verified by transformations and order change for summing probabilities.

Step-by-step solution

Write out the formulas for \(E(X)\) and \(F(x)\)

Recall that the expected value is calculated as \[E(X) = \sum_{x=0}^{\infty} x*p(x)\] and the cumulative distribution function is given by \[F(x) = \sum_{i=0}^{x} p(i)\]

Express \(1-F(x)\) in terms of \(p(x)\)

We can write \(1-F(x)\) as the sum of the probabilities \(p(x)\) from \(x+1\) to \(\infty\). This is because \(1-F(x)\) equals to the probability that \(X\) will take on a value greater than \(x\), which can be represented by the sum of all probabilities from \(x+1\) to \(\infty\): \[1 - F(x) = \sum_{i=x+1}^{\infty} p(i)\]

Substitute into the equation

We can substitute the equation from step 2 into the given equation of the problem: \[\sum_{x=0}^{\infty}[1-F(x)] = \sum_{x=0}^{\infty} \left(\sum_{i=x+1}^{\infty} p(i)\right)\] Note that for each inner sum, \(i\) starts from \(x+1\). But we still sum over all possible \(x\) from 0 to \(\infty\) in the outer sum.

Change the order of summation

Now, we are going to change the order of the summation to simplify the calculation. Observe that for each \(i\), it will be added once for each \(x\) starting from 0 up to \(i-1\), so: \[\sum_{x=0}^{\infty} \left(\sum_{i=x+1}^{\infty} p(i)\right) = \sum_{i=1}^{\infty} \left(\sum_{x=0}^{i-1} p(i)\right)\] which further simplifies to \( \sum_{i=1}^{\infty} i*p(i)\]

Verify the final equation

Observe that the final equation \( \sum_{i=1}^{\infty} i*p(i)\) is exactly the definition of the expected value \(E(X)\) that we cited in step 1. Hence, the given equation is proved.