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This question already has an answer here:

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 $y>1$

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marked as duplicate by ncmathsadist, Clarinetist, drhab probability May 23 '18 at 20:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Have a look here or here. $\endgroup$ – drhab May 23 '18 at 19:52
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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: $$f_Y(y) = F_Y'(y) = \frac32y^{-5/2}$$

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

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