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Let $X$ be a standard exponential random variable, and $Y = \log X$.

(a) Find DIRECTLY the c.d.f of $Y$ and use it to calculate the density of $Y$.

(b) Find DIRECTLY the p.d.f of $Y$.

So far, I did:

(a) $F_Y(y) = P(\log X \leq y) = P(X \leq e^y) = F_X(e^y) f_Y(y) = F'_Y(y) = F'_X(e^y) = [f_X(e^y)][e^y] = (e^{-e^y})(e^y) = e^{y-e^y}$

(b) $f_Y(y) = P(\log X = y) = P(X = e^y) = f_X(e^y) = e^{-e^y}$

How come the answers I got for part a and b are not the same. What did I do wrong?

Mikasa
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woaini
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1 Answers1

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Because PDFs don't work like this. The issue is you need to change of variables on the probability $P(Y=y)=f_Y(y)dy$, and not just the density. More precisely, if $y=\log(x)$ then $dy=e^{-y}dx$, so $P(Y=y)=f_Y(y)dy=P(\log X=y)=f_X(e^y)dx=f_X(e^y)e^ydy$ so that you recover your answer from the first part.

Alex R.
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  • Sorry but the identity $P(Y=y)=f_Y(y)dy$ is a monstrosity, which should be avoided at all cost. – Did Oct 22 '12 at 05:43
  • If y = log x, then shouldn't dy = (1/x) dx? – woaini Oct 22 '12 at 06:04
  • @woaini $y=\log x\implies e^y=x\implies e^ydy=dx\implies dy=e^{-y}dx$. Both are correct, since $\frac 1x=e^{-y}$. – Robert Mastragostino Oct 22 '12 at 06:20
  • Just for future reference I believe that you need to use the pdf $f_{X>0}$ since $Y=\log(X)$ is only defined when $X>0$. So $f_Y(y)=e^yf_{X>0}(e^y)=2e^2f_X(e^y)$. This pdf will integrate to $1$ over $\mathbb R$ however $f_X(e^y)e^y$ will only integrate to $\frac{1}{2}$ over $\mathbb R$. You could also consider $Y=\log|X|$ and here we would avoid needing to condition $X>0$ and we would again see that $f_Y(y)=2e^yf_X(e^y)$ – alpastor Apr 12 '19 at 15:24