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If Xis an exponential random variable with mean 1/λ, show that

EXk=k!λkk=1,2,…

Hint: Make use of the gamma density function to evaluate the preceding.

Short Answer

Expert verified

The statement has been proved true, i.e.

E(Xk)=k!λk;k=1,2,...

Step by step solution

01

Step 1. Defining X

The probability density function of random variable X is

f(x)=1λe−x/λ;x>0

02

Step 2. Calculations

Transforming the above variable to Gamma distribution and finding the kth raw moment, we get-

E(Xk)=1Γ(t)∫0∞xkλe−λx(λx)t−1dx=λ−kΓ(t)∫0∞λe−λx(λx)t+k−1dx=λ−kΓ(t)Γ(t+k);k=1,2,...

Putting t=1yields an exponential distribution, therefore, the final expression becomes

=λ−kΓ(1)Γ(t+1)E(Xk)=k!λk;k=1,2,...

which proves the required expression.

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