Transforming functions of random variables (exponential->Weibull)

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SUMMARY

The discussion centers on transforming a random variable from an exponential distribution to a Weibull distribution. Given that X follows an exponential distribution with parameter L, the transformation Y = X^(1/a) leads to the density function fy(s) = La(s^[a-1])e^(-L[s^a]). Despite the calculations appearing correct, the user questions whether this result is a variation of the Weibull distribution. The consensus is that the transformation is valid and yields a Weibull-like distribution.

PREREQUISITES
  • Understanding of exponential distributions and their properties
  • Familiarity with the Weibull distribution and its applications
  • Knowledge of probability density functions (PDFs) and cumulative distribution functions (CDFs)
  • Basic skills in mathematical transformations of random variables
NEXT STEPS
  • Study the properties of the Weibull distribution in detail
  • Explore the derivation of the Weibull distribution from other distributions
  • Learn about the applications of Weibull distributions in reliability engineering
  • Investigate the implications of parameter 'a' in the transformation Y = X^(1/a)
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Statisticians, data scientists, and students in probability theory who are interested in understanding transformations of random variables and their implications in statistical modeling.

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Homework Statement



Suppose X has an exponential with parameter L and Y=X^(1/a).
Find the density function of Y. This is the Weibull distribution

Homework Equations





The Attempt at a Solution



X~exponential (L) => fx(s)= Le^(-Ls)

Fx(s)=P(X<s) = 1-e^(-Ls)

P(Y<s)=P(X^(1/a)<s)=P(X<s^a)= 1-e^(-L[s^a])= Fy(s)

thus fy(s) = La(s^[a-1])e^(-L[s^a))

However, this doesn't seem to be the Weibull distribution. Did I do something wrong? Or is this just a variation of it?
 
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Looks fine to me.
 
Alright, thanks!
 

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