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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}

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  • $\begingroup$ 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
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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 ....$$\begin{split}\mathsf P(Y\leq y)&=\mathsf P(\mathrm e^{X/2}\leq y)\\[1ex]&=\mathsf P(X\leq 2\ln y)\\[2ex]F_Y(y)&=F_X(2\ln y)\\[2ex]f_Y(y)&=\begin{vmatrix}\dfrac{\mathrm d F_X(2\ln y)}{\mathrm d y\qquad\quad}\end{vmatrix}\\[1ex]&=\begin{vmatrix}\dfrac{\mathrm d (2\ln y)}{\mathrm d y\qquad~~~}\end{vmatrix}f_X(2\ln y)\\&~~\vdots\end{split}$$

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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