Chapter 5: Q. 5.19 (page 216)

If $X$ is an exponential random variable with mean $1 / λ$ , show that
$E [X^{k}] = \frac{k!}{λ^{k}} k = 1, 2, \dots$
Hint: Make use of the gamma density function to evaluate the preceding.

Short Answer

Expert verified

The statement has been proved true, i.e.

$E (X^{k}) = \frac{k!}{λ^{k}}; k = 1, 2, ...$

Step by step solution

Step 1. Defining X

The probability density function of random variable X is

$f (x) = \frac{1}{λ} e^{- x / λ}; x > 0$

Step 2. Calculations

Transforming the above variable to Gamma distribution and finding the k^th raw moment, we get-

$\begin{matrix} E (X^{k}) = \frac{1}{Γ (t)} \int_{0}^{\infty} x^{k} λ e^{- λ x} {(λ x)}^{t - 1} d x \\ = \frac{λ^{- k}}{Γ (t)} \int_{0}^{\infty} λ e^{- λ x} {(λ x)}^{t + k - 1} d x \\ = \frac{λ^{- k}}{Γ (t)} Γ (t + k); k = 1, 2, ... \end{matrix}$

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

$\begin{matrix} = \frac{λ^{- k}}{Γ (1)} Γ (t + 1) \\ E (X^{k}) = \frac{k!}{λ^{k}}; k = 1, 2, ... \end{matrix}$

which proves the required expression.

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If $X$ is an exponential random variable with mean $1 / λ$ , show that
$E [X^{k}] = \frac{k!}{λ^{k}} k = 1, 2, \dots$
Hint: Make use of the gamma density function to evaluate the preceding.

Short Answer

Step by step solution

Step 1. Defining X

Step 2. Calculations

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If Xis an exponential random variable with mean 1/λ, show thatEXk=k!λkk=1,2,…Hint: Make use of the gamma density function to evaluate the preceding.

Short Answer

Step by step solution

Step 1. Defining X

Step 2. Calculations

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Statistics

Geometry

Probability and Statistics

Calculus

Pure Maths

Study anywhere. Anytime. Across all devices.

If $X$ is an exponential random variable with mean $1 / λ$ , show that
$E [X^{k}] = \frac{k!}{λ^{k}} k = 1, 2, \dots$
Hint: Make use of the gamma density function to evaluate the preceding.