Chapter 5: Q. 5.25 (page 216)

Let $Y = \frac{X - 𝜈}{𝛼}$ .
Show that if X is a Weibull random variable with parameters ν, α, and β, then Y is an exponential random variable with parameter λ = 1 and vice versa.

Short Answer

Expert verified

The above statement is proved.

Step by step solution

Given information

We are assuming thatXis a Weibull with parameters $𝜈, 𝛼 and 𝛽$

Calculation

We have

$P \{{(\frac{X - 𝜈}{𝛼})}^{𝛽} \leq x\} = P \{(\frac{X - 𝜈}{𝛼}) \leq x^{\frac{1}{𝛽}}\} = P (X \leq 𝜈 + 𝛼 x^{\frac{1}{𝛽}}) = 1 - e^{- x}$

which is nothing but the CDF of exponential distribution.

Therefore it's proved that Ywill follow exponential distribution with parameter $𝜆 = 1$

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Let $Y = \frac{X - 𝜈}{𝛼}$ .
Show that if X is a Weibull random variable with parameters ν, α, and β, then Y is an exponential random variable with parameter λ = 1 and vice versa.

Short Answer

Step by step solution

Given information

Calculation

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Let Y=X-𝜈𝛼.Show that if X is a Weibull random variable with parameters ν, α, and β, then Y is an exponential random variable with parameter λ = 1 and vice versa.

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Statistics

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Study anywhere. Anytime. Across all devices.

Let $Y = \frac{X - 𝜈}{𝛼}$ .
Show that if X is a Weibull random variable with parameters ν, α, and β, then Y is an exponential random variable with parameter λ = 1 and vice versa.