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Let $X$ be an exponential random variable with parameter $\lambda=9$. Let $Y$ be the random variable defined by $Y=10e^X$. Compute the probability density function of $Y$: what is $f_Y(t)$ (for $t\geq10)$?

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    $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Mar 15 '18 at 20:01
  • $\begingroup$ I assume you mean $Y=10e^X$. Please make an effort to compute $P[Y\leq y]$. $\endgroup$ – Michael Mar 15 '18 at 20:26
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Here are some hints to get you started.

The cdf of $Y$ is given by

$$\begin{align*} F_Y(t) &=P(Y\leq t)\\\\ &=P(10e^X \leq t) \end{align*}$$

Can you go from here to get this in terms of the cdf of $X$? That is, $P(X\leq x)$.

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  • $\begingroup$ I calculated for a few times but I am not sure whether this is correct: 10e^x <= t ln10+x <= lnt x <= lnt-ln10 $\endgroup$ – Qihong Dai Mar 16 '18 at 0:31
  • $\begingroup$ Please include your efforts into your original question. $\endgroup$ – Remy Mar 16 '18 at 0:33
  • $\begingroup$ calculated for a few times but I am not sure whether this is correct: 10e^x <= t ln10 + x <= lnt x <= lnt - ln10 Then cdf of X: e^(-9x)dx | (lny/10 ~ 0) = (-1/9)e^(-9x) | (lny/10 ~ 0) Could you tell me what should I do next? $\endgroup$ – Qihong Dai Mar 16 '18 at 0:40
  • $\begingroup$ Please edit this into your original question and use mathjax to format it to improve readability. $\endgroup$ – Remy Mar 16 '18 at 0:47

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