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Let X be a random variable which follows an exponential distribution with parameter $\lambda$ ($\lambda>0$), find the distribution of the random variable $Y = −3\ln(X)$.

So this is my answer for that

$P(Y\leq y) = P(−3lnX \leq y) = P(X\geq e^{-\frac{y}{3}}) = 1-F_x(e^{-\frac{y}{3}})$

$f_Y(y) = \frac{\lambda}{3} e^{-\frac{y}{3}} e^{-\lambda e^{-\frac{y}{3}}}, -\infty \leq y \leq \infty$

We have this definition

If T follows an Gumbel distribution with parameters $\alpha$ and $\beta$, with Probability Density Function $$f_T(t)=\frac{e^{-(t-\alpha)/\beta}}{\beta}e^{-e^{-(t-\alpha)/\beta}}, -\infty \leq t \leq \infty$$

So Y follows Gumbel distribution with parameters $\alpha=0$ e $\beta=3$?

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  • $\begingroup$ This is correct if $\lambda =1$. $\endgroup$ – Kavi Rama Murthy Apr 12 at 5:46
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Your PDF: $f_Y(y)=\frac{\lambda}{3} e^{-y/3}e^{-\lambda e^{−y/3}}$. All good

Target PDF: $f_T(y) = \frac{e^{-(y-\alpha))/\beta}}{\beta}e^{-e^{-(y-\alpha)/\beta}}$.

So we would like $\lambda e^{-y/3} = e^{-(y-\alpha)/\beta}$.

How do we do that? Taking logs (just for exploratory purposes)

$\ln (\lambda) - y/3 = -y/\beta + \alpha/\beta$

So we will want $\beta = 3$, $\alpha = 3 \ln (\lambda)$.

It also follows that $e^{-(y-\alpha)/\beta}/\beta = \lambda/3 e^{-y/3}$. So it all works!

Hence you can fit a Gumbel to it regardless of the value of $\lambda$ in your exponential distribution.

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