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Let $X$ be a random variable which follows an exponential distribution with parameter $\lambda (\lambda > 1)$. Define a new random variable $Y$ by $$Y = e^X$$ Find the CDF of $Y$.

So is this saying that I need to find the CDF of $\lambda e^{-\lambda x}$ or $e^{\lambda e^{-\lambda x}}$?

Edit: This is the answer I came up with although not entirely sure if this is correct.

$$F_y(y) = P(e^X \leq y) = P(X \leq ln(y)) = F_x(ln(y))$$ $$ F_y(y) = \begin{cases} 1-e^{-\lambda ln(y)} & y \geq 0 \\ 0, & otherwise \end{cases} $$ $$f_y(y) = \frac{1}{y} f_x(ln(y))$$

$$E[Y] = \int_{1}^{\infty} f_x(ln(y)) dy$$

Feel free to correct me if I'm wrong.

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  • $\begingroup$ you will find the CDF of the random variable $Y=e^X$, where $X$ is exponentially distributed. So your question in the last line makes no sense. $\endgroup$ – Seyhmus Güngören May 23 '17 at 23:31
  • $\begingroup$ The constant confusion between $X$ and $Y$ vs $x$ and $y$, does not help your understanding. A first step would be to reach some amount of rigor in this respect. $\endgroup$ – Did May 27 '17 at 9:28
  • $\begingroup$ Please enlighten me then. $\endgroup$ – saechaoc May 27 '17 at 17:41
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Neither $\lambda e^{-\lambda x}$ nor $e^{\lambda e^{-\lambda x}}$ is $Y$.

$Y$ is $e^X$, so find the CDF of $Y$ means compute $P(Y\leq y) = P(e^X\leq y)$.

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  • $\begingroup$ Ok thank you! The confusion I had was since X is exponentially distributed I assumed it was $\lambda e ^{-\lambda x}$ but that makes more sense. $\endgroup$ – saechaoc May 24 '17 at 0:09
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Let

$$f(x) = \lambda e^{-\lambda x}$$

with domain of support given by $\{x,\ 0,\ \infty \}$.

The transformed random variable,

$$Y = \exp(X)$$

gives the new distribution

$$f(y) = \frac{1}{y^{1+\lambda}}$$

with domain of support $\{y,\ 1,\ \infty \}$

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