Find the pdf of Y = $e^{\frac{1}{2}x}$ if x is exponentially distributed with parameter $\lambda =$ 3.

This question was given to me as a review for an upcoming exam, I am not really sure where to start.

I'm thinking :

$F_Y(y) =$

\begin{cases} \lambda e^{-\lambda x} & x\geq 0 \\ 0 & otherwise \end{cases}

as a start.

Update Attempt:

$P(Y \leq y)$

$= e^{\frac{X}{2}} \leq y$

$X \leq 2ln(y)$

$\frac{d}{dy}(1-e^{-2 \lambda ln(y)})$

$= \frac {2 \lambda e^{-2\lambda ln(y)}}{y}$

$F_Y(y)=$ \begin{cases} \frac {2 \lambda e^{-2\lambda ln(y)}}{y} & y\geq 1 \\ 0 & otherwise \end{cases}$F_y(Y)=$ \begin{cases} \frac {2 (3) e^{-2(3) ln(y)}}{y} & y\geq 1 \\ 0 & otherwise \end{cases} $F_Y(y)=$ \begin{cases} \frac {6 e^{-6 ln(y)}}{y} & y\geq 1 \\ 0 & otherwise \end{cases}

$F_Y(y)=$ \begin{cases} 6y^{-7} & y\geq 1 \\ 0 & otherwise \end{cases}

  • $\begingroup$ Have you learnt a "change of variables" formula for PDF's? If not, you can also do this by finding the CDF of $Y$ (which is $P(Y\le y)$) and then differentiating it with respect to $y$. $\endgroup$ – Minus One-Twelfth Mar 20 at 2:56
  • $\begingroup$ I have not learned of change of variables formula for PDF's as far as I know. I will work with the hint you provided and update when I can. $\endgroup$ – Joe Mar 20 at 2:58
  • $\begingroup$ I've added an edit with a few more attempted steps. $\endgroup$ – Joe Mar 20 at 3:05
  • $\begingroup$ You can't say $\mathsf P(Y\leq y)=1-\mathrm e^{-\lambda y}$ because you do not know that $Y$ is exponentially distributed (in fact, it is not). You only know the distribution for $X$, and that $Y=\mathrm e^{X/2}$. $\endgroup$ – Graham Kemp Mar 20 at 3:27
  • $\begingroup$ The letters are case sensitive. Capital and lowers mean different things. Try harder to not conflate them. $\endgroup$ – Graham Kemp Mar 20 at 4:15

The first thing you have to do is master case sensitivity, otherwise you are just going to confuse yourself.   Don't mix up when you use random variable $Y$ and term $y$.

Also, your final answer should not contain any term $x$.   $F_Y(y)$ will clearly be a function of term $y$.

But moving on...

If $X$ is exponential with parameter $\lambda:=3$, and $Y=\mathrm e^{X/2}$, then the pdf for $Y$ is supported over $[1;\infty)$ and ....$$\begin{split}\mathsf P(Y\leq y)&=\mathsf P(\mathrm e^{X/2}\leq y)\\[1ex]&=\mathsf P(X\leq 2\ln y)\\[2ex]F_Y(y)&=F_X(2\ln y)\\[2ex]f_Y(y)&=\begin{vmatrix}\dfrac{\mathrm d F_X(2\ln y)}{\mathrm d y\qquad\quad}\end{vmatrix}\\[1ex]&=\begin{vmatrix}\dfrac{\mathrm d (2\ln y)}{\mathrm d y\qquad~~~}\end{vmatrix}f_X(2\ln y)\\&~~\vdots\end{split}$$

  • $\begingroup$ Could you explain how to get the domain [1,$\infty$)? $\endgroup$ – Joe Mar 20 at 3:34
  • $\begingroup$ @Joe It is just that the support for $X$ is $[0;\infty)$, so therefore that for $\mathrm e^{X/2}$ is $[1;\infty)$ $\endgroup$ – Graham Kemp Mar 20 at 3:37
  • $\begingroup$ Thanks, I somehow was completely oblivious to that. I understand that part now. I will add an update shortly with an attempt #2. $\endgroup$ – Joe Mar 20 at 3:38
  • $\begingroup$ I've updated my "Updated Attempt" section of the OP. $\endgroup$ – Joe Mar 20 at 3:49
  • $\begingroup$ Your formatting still needs work, but your working is correct. Notice that it can be simplified just a bit more, since $\mathrm e^{\ln y}=y$. $$f_Y(y)=\begin{cases} 6 y^{-7}&:& y\geqslant 1\\0&:& \textrm{otherwise}\end{cases}$$ $\endgroup$ – Graham Kemp Mar 20 at 4:24

From your attempt, you have written $P(Y\le y)= 1-e^{-\lambda y}$. This would be true if $Y$ were $Expo(\lambda)$, but we are not told $Y$ is (we are told $X$ is). Instead, note that for $y> 0$, we have $$Y\le y \iff e^{X/2}\le y\iff X \le 2\ln y.$$

Thus $P(Y\le y) = P(X\le 2\ln y)$. See if you can finish from here.

  • $\begingroup$ I've updated my "Updated Attempt" section of the OP. $\endgroup$ – Joe Mar 20 at 3:49
  • $\begingroup$ Note that (provided $2\ln y\ge 0$, i.e. $y\ge 1$), we have $\color{blue}{P(X\le 2\ln y) = 1- e^{-\lambda (2\ln y)}}$. So this is what we should differentiate. $\endgroup$ – Minus One-Twelfth Mar 20 at 3:53
  • $\begingroup$ So after I differentiate that, I use that in the pdf and then plug in the given $\lambda = 3$ parameter? (I updated the attempt to reflect my thought) $\endgroup$ – Joe Mar 20 at 4:01

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.