Question

Let \(X\) be a nonnegative random variable with cumulative distribution funetion \(F(x)=\operatorname{Pr}\\{X \leq x\\}\). Show $$ E[X]=\int_{0}^{\infty}[1-F(x)] d x $$ Hint: Write \(E[X]=\int^{\infty} x d F(x)=\int^{\infty}\left(\int_{0}^{x} d y\right) d F(x)\).

Short answer

To solve this proof, the Cumulative Distribution Function and Expected Value were broken down. The expression of expected value was then presented as a sum of two integrals. After interchanging the order of integration, the proof was made by equating the resultant expression with the original equation. Finally, it was shown that \(E[X] = \int_0^{\infty}[1 - F(x)] dx\).

Step-by-step solution

Understanding Cumulative Distribution Function

Firstly, you need to understand what a cumulative distribution function (CDF) is. The CDF, denoted as \(F(x)\), of a random variable is defined as the probability that the variable will take a value less than or equal to \(x\). In mathematical terms, \(F(x) = Pr\{X \leq x\}\). So, when the value of the CDF increases, the probability of the variable taking a value less than \(x\) also increases.

Understanding Expected Value

The expected value is the long-run average or mean value of random variables. It is calculated by integrating the product of the variable density and the variable value over the variable's range. In this case, the expected value of \(X\) is defined as \(E[X] = \int^{\infty} x d F(x)\).

Manipulating the Integrals

The hint provided suggests writing \(E[X]\) in terms of two integrals, i.e., \(E[X] = \int^{\infty}\left( ∫_{0}^{x} dy\right) dF(x)\). The inner integral can be interpreted as the area under the curve of the density function. Further, \(dF(x)\) is nothing but \(F'(x)dx\), which is the differential of the cumulative distribution function.

Interchanging the Integrals

The purpose of writing the integral in the nested form in step 3 is to allow the order of integration to be switched. Switching the order of integration is an important technique used in mathematics for simplifying complex integrals. By interchanging the order of integration, we rewrite \(E[X] = \int_0^\infty (1 - F(y)) dy\). From this equation, the expression \(1 - F(y)\) can be interpreted as the probability that the random variable \(X\) is greater than \(y\). Hence, it is evident that the expected value \(E[X]\) is equal to the sum of the probabilities that the random variable \(X\) exceeds each value \(y\) in its range.

Final Proof

Adding up the above steps, it is evident that the original equation can be re-expressed as: \(E[X] = \int_0^{\infty}[1 - F(x)] dx\), which is the required proof.