Question

If $X$is uniformly distributed over$(0,1)$, find the density function of$Y=e^X$.

Short answer

Therefore, the PDF of$Y=e^{X}is{f}_Y\left(y\right)=\left\{\begin{array}{ll}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$y$}\right.& if1<y<e\\ 0& o/w\end{array}\right.$

Step-by-step solution

Given information:

If$X$is uniformly distributed over$(0,1)$

Explanation:

First, we start with the cumulative density function$F_Y\left(y\right)$, since we want to find the PDF for$Y$. We haverole="math" localid="1646662591120" $F_Y\left(y\right)=P\left(Y\le y\right)=P\left(e^X\le y\right)=P\left(X\le \ln \left(y\right)\right)=F_X\left(\ln \left(y\right)\right)$, which is the CDF with respect to$X$.

Explanation:

After taking the derivative to get the PDF we obtain $f_Y\left(y\right)=\frac{d}{dy}{F}_X\left(\ln \left(y\right)\right)=F_X\left(\ln \left(y\right)\right)\times \frac{1}{y}$by the chain rule. But since$X~UNIF(0,1)$, we know that$f_X\left(\ln \left(y\right)\right)=1$and the domain changes to$e^0<e^X<e^1=1<y<e$, so we have that the PDF of$Y=e^{X}is{f}_Y\left(y\right)=\left\{\begin{array}{ll}\frac{1}{y}& if1<y<e\\ 0& o/w\end{array}\right.$