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I am trying to solve a problem which is quite easy but I just can't get the needed result.

Task: $x$~${Exp}(1)$. Therefore, pdf: $f_x(x)=e^{-x}$ and $y=lnx$.

I need to show that pdf of ${y=f_y(y)}=Exp(y-e^{y})$, y continues from minus infinity to plus infinity (standard extreme value type I distribution).

Formula in use: $$f_Y(y)=f_x(g^{-1}(y))\frac{1}{\lvert\frac{dy}{dx}\rvert}$$

Solving the problem gives me: $f_Y(y)=f_x(e^{y})\frac{1}{\lvert\frac{dy}{dx}\rvert}=(e^{-e^{y}}x)$

If I plug in the $x=e^{y}$ then I get the right result, but can I do that? Or maybe there is another way of achieving the ${y=f_y(y)}=Exp(y-e^{y})$ result?

lovetimberland
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1 Answers1

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Ok so $$F_Y(y) = P(Y \leq y) = P(ln (x) \leq y) = \int_{0}^{e^y} f_x(x) dx= \int_{0}^{e^y} e^{-x} dx = -e^{-e^{y}} + 1$$ Then $$f_Y(y)=F'_Y(y) = e^{y}e^{-e^y} = e^{y - e^y}$$

Ahmad Bazzi
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