Extreme Value Distribution

Question

I would appreciate a lead\tip on the next one:

Z is a standard random variable from the extreme value distribution. I need to show that $Y=\sigma \cdot Z+\mu$ is an extreme value variable with parameters - $\sigma$ and $\mu$.

Answer 1

Assuming $\sigma > 0$, you could start with finding an expression for the $Y$'s cumulative distribution function: $$ F_Y(y) = \Pr\left(Y \leqslant y\right) = \Pr\left(\sigma Z + \mu \leqslant y\right) = \Pr\left(Z \leqslant \frac{y-\mu}{\sigma}\right) = F_Z\left(\frac{y-\mu}{\sigma}\right) $$ By definition of the probability density function: $$ f_Y(y) = F_Z^\prime(y) = \frac{\mathrm{d}}{\mathrm{d}y} F_Z\left(\frac{y-\mu}{\sigma}\right) = f_Z\left(\frac{y-\mu}{\sigma}\right) \frac{1}{\sigma} $$ Now use definition of $f_Z(z)$ and compare the so obtained density function $f_Y(y)$ with that of the extreme value distribution with parameters $\mu$ and $\sigma$.