Let $X$ be a $\operatorname{Uniform}(0,1)$ random variable, and let $Y=e^{-X}$.

Find the CDF of $Y$.

Find the PDF of $Y$.

Find $\mathbb E[Y]$.

My problem

If I solve for the range of $y$ I get $\left(1, \frac 1e \right)$, but because $Y$ is not an increasing function, my second bound is smaller than my first. I am really confused as to how I would be able to solve for the CDF and PDF in this case... Any help would be greatly appreciated.


For $y$ in the interval $(1/e,1)$ we have $$\small F_Y(y)=\Pr(Y\le y)=\Pr(e^{-X}\le y)=\Pr(-X\le \ln y)=\Pr(X\ge -\ln y)=1-(-\ln y)=1+\ln y.$$ For completeness, we add that $F_Y(y)=0$ if $y\le 1/e$, and $F_Y(y)=1$ for $y\ge 1$. Note, in the above calculation, the switch in the direction of the inequality, when $\Pr(-X\le \ln y)$ was replaced by $\Pr(X\ge -\ln y)$.

Now the pdf is easy to find. For $E(Y)$, we can use the newly found pdf of $Y$, or else bypass the distribution of $Y$ entirely, computing $\int_0^1 e^{-x}\,dx$.

  • $\begingroup$ This is very helpful. Thanks. However, would the CDF be 0 for y < 1 and 1 for y > 1/e? This part does not really make sense. $\endgroup$ – Jonathan Yoder Oct 6 '15 at 21:33
  • $\begingroup$ Yes, I computed the cdf only for "interesting" $y$. But the cdf is defined for all $y$. If $y\gt 1$, then for $Y\le y$, so $F_Y(y)=\Pr(Y\le y)=1$. And if $y\lt 1/e$, we cannot have $Y\le y$, so $F_Y(y)=0$. $\endgroup$ – André Nicolas Oct 6 '15 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.