Question

If $X$is an exponential random variable with a parameter$\lambda =1$, compute the probability density function of the random variable $Y$defined by $Y=\log \left(X\right)$

Short answer

Therefore, the probability density function of the random variable$Y=e^y\times e^{-e^y}$

Step-by-step solution

Given information:

If X is an exponential random variable with parameter$\lambda =1$,

Explanation:

$F_x\left(x\right)=1-e^{-x}$

So if

$Y=\ln \left(x\right)$

We have

$$\begin{gathered} F_Y\left(y\right)=P(Y\le y) \\ =P(\ln \left(X\right)\le y) \\ =P(X\le e^y) \\ =F_X{e}^y \\ =1-e^{-e^y} \end{gathered}$$

Explanation:

Therefore, the probability density function of the random variable is

$$\begin{gathered} f_Y\left(y\right)=\frac{d}{dy}\left[1-e^{-e^y}\right] \\ =-\left(-e^y\right)\left(e^{-e^y}\right) \\ =e^y\times e^{-e^y} \end{gathered}$$