Alternative Approach to: $X,Y$ i.i.d, $\ln(X) \sim$Exp$(1)$. Find CDF of $XY$.

Question

This question was already answered here (I've also added an approach using $\ln(XY) = \ln(X) + \ln(Y)$ there) but I am asking this question to clear some things up about the other answer and ask whether my alternative albeit similar approach is correct or how to fix it.

Question First, @drhab states that “Apparently $X$ is a positive rv (if not then $\ln(X)$ would not be well defined).” But if $\ln(X) \sim$Exp$(1)$, this implies that $\ln(X) \ge 0$ almost surely and therefore $X \ge 1$ almost surely, which suffices to show that $\ln(X)$ is well defined, right?

Approach Since $\ln(X) \sim$Exp$(1)$, for all $k \ge 0$ we have \begin{equation*} \mathbb{P}(X \le e^k) = \mathbb{P}(\ln(X) \le k) = 1 - e^{-k} \end{equation*} By substitution I obtained \begin{equation*} \mathbb{P}(X \le a) = 1 - \frac{1}{a}. \end{equation*} for all $a \ge 1$. We can now obtain the density function by calculating the derivative: $f_X(x) = \frac{1}{x^2}$.

Now, for $z \ge 1$ we have \begin{align*} \mathbb{P}(XY \le z) & = \int_{1}^{\infty} 1_{(-\infty,z]}(x,y) \ \text{d}\mathbb{P}(x,y) = \int_{1}^{\infty} \int_{\mathbb{R}^+} 1_{\left(1, \frac{z}{y}\right)}(x) \ \text{d}\mathbb{P}_{X} \ \text{d}\mathbb{P}_{Y} \\ & = \int_{1}^{z} \int_{\mathbb{R}^+} 1_{\left(1, \frac{z}{y}\right)}(x) \frac{1}{x^2} \frac{1}{y^2} \ \text{d}x \ \text{d}y \\ & = \int_{1}^{z} \frac{1}{y^2} \left[ \int_{1}^{\frac{z}{y}} \frac{1}{x^2} \ \text{d}x \right] \text{d}y \\ & = \int_{1}^{z} \frac{1}{y^2} \left(1 - \frac{y}{z}\right) \text{d}y = 1 - \frac{1 + \ln(z)}{z}. \end{align*}

Answer 1

As @stochasticboy321 correctly pointed out, the mistake I made was with the upper bound of my integral.

As he mentioned as well, one could also notice that the sum $\ln(XY) = \ln(X) + \ln(Y)$ of two exponentially distributed random variables is $\Gamma(2,1)$-distributed, i.e. the CDF of $\ln(Z)$ is $$ e^{-1x} \sum_{k = 2}^{\infty} \frac{(1x)^k}{k!} = 1 -e^{-x} x - e^{-x} $$ This means for $z \ge 1$ $$ \mathbb{P}(Z \le e^z) = \mathbb{P}(\ln(Z) \le z) = 1 - \frac{z + 1}{e^{z}} \implies \mathbb{P}(Z \le z) = 1 - \frac{\ln(z) + 1}{z} $$

Answer 2

To address your first question, I would phrase the problem in a more precise way to make it clear everything is well-defined. Let $Z,Z'$ be iid random variables such that $$\mathbb P(Z>t)=e^{-t},\quad \forall t\geq 0.\qquad (1)$$ Let $X=e^Z$ and let $Y=e^{Z'}$. (There is no question that everything is well-defined in this formulation, and that it is equivalent to yours.)

Translating your calculations into a more probabilistic language, $$ \mathbb P(XY\leq z\mid Y)=\mathbb P\bigl(X\leq \frac{z}{Y}\mid Y\bigr)=\bigl(1-\frac{Y}{z}\bigr)1_{Y\leq z}, $$ where we have applied $(1)$ with $e^t=z/Y$ in the last equality. Then, $$ \mathbb P(XY\leq z)=\mathbb E\bigl(1-\frac{Y}{z};Y\leq z\bigr). $$ Let me now point out a slightly different way of performing this last integral, which perhaps makes it more transparent where the terms are coming from. $$ \mathbb E\bigl(1-\frac{Y}{z};Y\leq z\bigr)=\mathbb E\bigl(1-\frac{e^{Z}}{z};Z\leq \log z\bigr)=\int_0^{\log z}(1-\frac{e^x}{z})\cdot (e^{-x} dx).$$ Now the integrand equals $e^{-x}-1/z$. Integrating the first term gives $1-1/z$, and integrating the second gives $$-\frac{\log z}{z},$$ which yields the desired expression.