How to show that $ln \ Y$ of uniform distributed random variable is exponential distributed

Question

We have a random variable called $U$

$U \sim unif[0,1]$

$Y:=1/U^{1/3}$

Then I had to determine CDF, PDF and mean which are the following

(i) CDF

$F_{Y}(b)=\begin{cases} \frac{1}{b^3}\ \ \ b \in[0,1] \\ 0 \ \ \ \ \ b<0\\ 1 \ \ \ \ \ b>1 \end{cases}$

(ii) PDF

$f(b)=F'(B)=\begin{cases} -3b^{-4}\ \ \ \ b\in[0,1] \\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \text{else} \end{cases}$

(iii) mean

$\mathbb{E}[Y]=\mathbb{E}[1/U^{1/3}]=\int_{- \infty}^{\infty} \frac{1}{u^{1/3}}\cdot \frac{1}{1} \mathbb{1}_{[0,1]}(u) du=\int_0^1 \frac{1}{u^{1/3}} du=\left[\frac{3}{2}u^{2/3}\right]_0^1=\frac{3}{2}$

Now I have to show that $ln \ Y$ is exponential distributed with a parameter $\lambda$. But I don't how the approach for this problem looks like.

Answer 1

It's quite simple given you have derived the distribution of $Y$, but you have not done this correctly, the negative PDF should have been a tip off...

First, let's get the CDF of $Y$:

\begin{align*} F_Y(y) &= P(Y \le y)\\ & = P(U^{-1/3} \le y)\\ & = P(U^{1/3} \ge \frac{1}{y})\\ & = P(U \ge y^{-3})\\ & = 1- P(U \le y^{-3})\\ & = 1-F_U(y^{-3})\\ & = 1 - y^{-3} \end{align*}

Therefore, the pdf of $Y$ is $f_Y(y) = 3y^{-2}$

Now, let $Z = lnY$, then:

\begin{align*} F_Z(z) & = P(Z \le z) \\ & = P(lnY \le z)\\ & = P(Y \le e^z)\\ &= F_Y(e^z)\\ & = 1 - (e^z)^{-3}\\ & = 1-e^{-3z} \end{align*}

and so the pdf of $Z$ is:

$$ f_Z(z) = 3e^{-3 z}, \quad z > 0 $$

therefore, $Z \sim \text{expo}(\lambda =3)$ and so $E(Z) = \lambda^{-1} = 1/3$