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Let $X$ have a Weibull distribution with shape parameter $\alpha$ and scale parameter $\beta .$ Show that the transformed variable $Y=\ln (X)$ has an extreme value distribution as defined in Section $3.6,$ with $\theta_{1}=\ln (\beta)$ and $\theta_{2}=1 / \alpha$

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

Temple University

Missouri State University

Piedmont College

Cairn University

Lectures

02:58

Let $X$ have a Weibull dis…

04:26

05:39

Let $X$ and $Y$ be indepen…

02:05

Let$$Y=\left(\frac…

02:42

Show that a plot of $\log …

03:38

Consider a sample of $x_{1…

02:13

Derive the cdf for the Wei…

were given a Waibel distribution with Parameters Alfa and Beta. And we want to show that the transformation why equals the natural algorithm of X is an extreme value distribution with parameters Data one is equal to lawn of data and data to is equal to one over Alfa. Let's begin by solving for the pdf of why sorry. Let's we'll sell for the CDF of why oh so substituting in lawn of X for why this is equal to the probability that X is less than or equal to e to the Y. Now, this is the cumulative distribution for X evaluated at X is equal to eat the exponents. Why? So we can substitute E to the exponents? Why into this equation and it is equal. So now we will start working from the other end. So we will look at the CDF of the extreme value distribution and we will have data one equal to the natural log of B and theater to equal to one over Alfa. So they should all be in brackets right here. So I'm just going thio in this argument here I'm going to multiply the numerator and the denominator by Alfa So we end up with Alfa Zed minus Alfa times, a natural log rhythm of beta. And now I'm going to use the rule for exponents. So if you have a difference in the argument you can represent that is as follows and using rules for exponents. So we can take this factor, Alfa, and express it as an exponents over here. And this is one minus a to the negative. Alpha's ed guided by Better to the Alfa. This whole thing is the argument of the exponents. So this is the CDF for the extreme value distribution up here we have the CDF for why now, in terms of the domain for why? For the Waibel X is greater than or equal to zero. If why is the natural algorithm of X then why goes from negative infinity to infinity and therefore we can see that CDF of why is identical to the CDF of the extreme value distribution where the extreme if the extreme value distribution has the parameters, the other one is equal to the natural log of data and data to is equal to one over Alfa

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