# PDF of $Y=e^{2X}$ given X~exp(3) [duplicate]

Consider the random variable X∼exp(3). And let $Y=e^{2X}$.

Find the PDF of Y.

I know by properties of exp distribution that the PDF of $f_X(x)$ is given by $$f_X(x) = \begin{cases} 3e^{-3x}; & \text{ if, } x > 0 \\ 0; & \text{ otherwise } \end{cases}$$

The way I approched the problem is by first trying to find the CDF of Y and then taking the derivative to get the PDF of Y.

This is a exercise on a previous exam so I know the result and I'm getting a wrong one. I think I'm having troubles finding the CDF. This is what I've tried

$F_Y(y) = P(Y\le y)=P(e^{2X}\le y)=P(X\le \frac{ln(y)}{2})$

So I have the upper limit and the lower limit of the integral is 1, as $e^{2\cdot 0}$ is 1 and that is the lowest that $x$ gets.

I then integrate the the whole thing to find the CDF.

$\int_{1}^{\frac{ln(y}{2}} 3e^{-3X} \ dx$

But I'm doing something wrong, because it's suppose to give me the result of $\frac{3}{2}y^\frac{-5}{2}$, for $y>1$

## marked as duplicate by ncmathsadist, Clarinetist, drhab probability StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 23 '18 at 20:03

• Have a look here or here. – drhab May 23 '18 at 19:52

Actually $X$ can go to zero, so the correct lower bound for $x$ is $0$ instead of $1$, giving us: $$F_Y(y) = \int_{0}^{(\ln y)/2} 3e^{-3x} \ \mathrm dx = 1 - y^{-3/2}$$

Therefore: $$f_Y(y) = F_Y'(y) = \frac32y^{-5/2}$$

• Thanks for the reply. Can you explain why X can go to zero? – Sirmimer May 23 '18 at 19:58
• $X$ is exponentially distributed, so its pdf has a support over $[0;\infty)$ @Sirmimer – Graham Kemp Jun 30 at 6:59