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Thread: Expected Value of Exponential Distribution

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    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.

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    Quote Originally Posted by moberle View Post
    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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