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Chapter 5: Q. 5.19.TE (page 216)

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

E[Xk]=k!λkk=1,2,…

Short Answer

Expert verified

The statement has been proved true, i.e.

E(Xk)=k!λk;k=1,2,...

Step by step solution

01

Given information.

Xis an exponentially distributed random variable with a mean1λ

02

Step 2. Defining X.

The probability density function of a random variable X is

f(x)=1λe−x/λ;x>0

03

Step 3. Calculation.

Transforming the above variable to Gamma distribution and finding the kthraw 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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A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a specified number, then you either win 35 if the roulette ball lands on that number or lose 1 if it does not. If you continually make such bets, approximate the probability that

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