# If $X$ is an exponentially distributed variable with mean $\lambda$, $Y=−3\ln(X)$ has Gumbel distribution?

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 $$\beta=3$$?

• This is correct if $\lambda =1$. – Kavi Rama Murthy Apr 12 at 5:46

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.