Thread: Expected Value of Exponential Distribution

1. Expected Value of Exponential Distribution

I've almost completed this problem, but I'm stuck on the last part. The first two parts that I've already shown are also given:

a) For a whole numbered, non-negative random variable, show that it is true that:
E(X) = Sum(n=1 to infinity) of P(X>=n)

b)Let X be a non-negative continuously distributed random variable with probability density f and distribution function F. Show that:
E(X) = Integral(0 to infinity) of (1-F(X))dx

c) Here's the one that I can't get.
Formulate and prove an appropriate formula for E(X^n), n is a natural number. Use this formula to find the expected value of the Exponential Distribution with Parameter lambda.

I can get the second part of this question, but I need some help getting started by finding the formula for E(X^n). I'm not sure where to start. I know that the expected value of the Exponential Distribution is simply lambda. Any hints? Thanks.

2. Originally Posted by moberle
I've almost completed this problem, but I'm stuck on the last part. The first two parts that I've already shown are also given:

a) For a whole numbered, non-negative random variable, show that it is true that:
E(X) = Sum(n=1 to infinity) of P(X>=n)

b)Let X be a non-negative continuously distributed random variable with probability density f and distribution function F. Show that:
E(X) = Integral(0 to infinity) of (1-F(X))dx

c) Here's the one that I can't get.
Formulate and prove an appropriate formula for E(X^n), n is a natural number. Use this formula to find the expected value of the Exponential Distribution with Parameter lambda.

I can get the second part of this question, but I need some help getting started by finding the formula for E(X^n). I'm not sure where to start. I know that the expected value of the Exponential Distribution is simply lambda. Any hints? Thanks.
Expected value of exponential is 1/lambda if lambda is the "rate" parameter.
E(X^n)=integral 0 to infinity x^n * lambda exp(-lambda*x) dx
use the substitution t=lambda*x
The integral will re-evaluate to
(1/lambda^n)integral 0 to infinity t^n e^-t dt
=(1/lambda^n) gamma(n+1) (see definition of gamma below)
=n!/lambda^n
For expectation put n=1 to get answer 1/lambda

gamma(n)=integral 0 to infinity x^(n-1)*exp^-x dx=(n-1)! when n is an integer.

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