# Finding the PDF of a function if X is exponentially distributed with a given parameter $\lambda$ = 3

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:

$$P(Y \leq y)$$

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

$$X \leq 2ln(y)$$

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

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

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

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

• 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 $y$. – Minus One-Twelfth Mar 20 at 2:56
• 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. – Joe Mar 20 at 2:58
• I've added an edit with a few more attempted steps. – Joe Mar 20 at 3:05
• 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 $Y=\mathrm e^{X/2}$. – Graham Kemp Mar 20 at 3:27
• The letters are case sensitive. Capital and lowers mean different things. Try harder to not conflate them. – 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 ....$$\begin{split}\mathsf P(Y\leq y)&=\mathsf P(\mathrm e^{X/2}\leq y)\\[1ex]&=\mathsf P(X\leq 2\ln y)\\[2ex]F_Y(y)&=F_X(2\ln y)\\[2ex]f_Y(y)&=\begin{vmatrix}\dfrac{\mathrm d F_X(2\ln y)}{\mathrm d y\qquad\quad}\end{vmatrix}\\[1ex]&=\begin{vmatrix}\dfrac{\mathrm d (2\ln y)}{\mathrm d y\qquad~~~}\end{vmatrix}f_X(2\ln y)\\&~~\vdots\end{split}$$

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

• I've updated my "Updated Attempt" section of the OP. – Joe Mar 20 at 3:49
• 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. – Minus One-Twelfth Mar 20 at 3:53
• 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) – Joe Mar 20 at 4:01