how about a Weibull?
Weibull Probability Density Function
The Extreme Value Distribution has cdf of F(x) = 1 - exp( -e^((x-a)/b) ) for -inf < x < inf. Let Y = e^X. Find the cdf and pdf of Y. What is the distribution of Y?
So far, for the cdf of Y, I have F(y) = P(Y<y) = P(e^X < y) = P(X < ln(y)), and I plugged ln(y) into F(x) to get: 1 - e^(-e^((ln(y)-a)/b)) = 1 - e^((-y^(1/b))*(e^(-a/b))).
I know to get the pdf I'll just have to take the derivative of F(y) wrt y, but I'm not recognizing any familiar distribution with the cdf I have now. Did I make a mistake somewhere or can it be simplified further?