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Let $X \sim \operatorname{Unif}(0,1)$. Let $Y=r(X)=e^{X}$. Then, $$ \mathbb{E}(Y)=\int_{0}^{1} e^{x} f(x) d x=\int_{0}^{1} e^{x} d x=e-1 $$ Alternatively, you could find $f_{Y}(y)$ which turns out to be $f_{Y}(y)=1 / y$ for $1<y<e$. Then, $\mathbb{E}(Y)=\int_{1}^{e} y f(y) d y=e-1$.

Question: How does one obtain $1/y$ in computing the marginal mass function of $y$?

IamKnull
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critterTI
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    You can use a change of variables, or find $\mathbb P(Y \le y)$ and then differentiate – Henry Jun 14 '21 at 10:53

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Hint : $F_Y(y)=P(e^X \leq y)$ will give the CDF and its dericative will give the PDF (probability distribution function)

Try yourself and if you can't do it yourself then refer this

IamKnull
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