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Let $X$ be an exponential random variable with parameter $\lambda = 4$, and let $Y$ be the random variable defined by $Y=8e^X$. Compute the probability density function of $Y$:

$f_Y(t) = $

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  • $\begingroup$ Is there a technique you are familiar with? Method of transformations? Or alternately (my preference) first finding an expression for the cdf of $Y$, that is, for $\Pr(Y\le y)$. $\endgroup$ – André Nicolas Nov 9 '15 at 5:05
  • $\begingroup$ @AndréNicolas I believe I'm supposed to use the technique of finding an expression for the cdf of $Y$ since that is what we have most recently learned in class relating to the subject. I haven't seen an example where the random variable is defined in terms of another one so I'm not sure how to go about doing this. $\endgroup$ – user3370201 Nov 9 '15 at 5:07
  • $\begingroup$ The question has been answered. The answer can serve as a template in similar questions. $\endgroup$ – André Nicolas Nov 9 '15 at 5:18
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Hint: first find the cdf of $Y$ and use the cdf of $X$ to help with the computation. $$P(Y \le y)=?$$

$$P(Y \le y) = P(8e^X \le y) = P(X \le \log(y/8))$$

Once you have the cdf $P(Y \le y)$, take the derivative with respect to $y$ to get the pdf.

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  • $\begingroup$ This would give me a cdf of $\int \lambda e^{-\lambda ln(\frac{t}{8})} dt$ right? And then I would have to take the derivative of this to get the pdf? $\endgroup$ – user3370201 Nov 9 '15 at 5:14
  • $\begingroup$ Yes; note also that the cdf of an exponential random variable has a simpler form than what you have written. $\endgroup$ – angryavian Nov 9 '15 at 7:02

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