Density of product of uniform random variable and exponential random variable

Question

I was trying to find density $\rho_{\xi\eta}$ if $\xi \sim U(0,1)$ and $\eta \sim Exp(\lambda)$ and they are independent.

Using convolution of product, I got

$\rho_{\xi\eta}(v) = \mathbb{1}_{\{v\geq0\}}\int_{0}^{1}\lambda e^{-\lambda v u^{-1}}u^{-1}du$, there $\mathbb{1}$ is and indicator function.

But as I know this is the Exponential integral, so It can be evaluated in elementary functions. Did I do something wrong and I need to find mistakes in calculations or is it the right answer?

Here is my calculations:

As I said, $\mathbb{1}_{x \in A}(x)$ is the Indicator function and it equals $1$ if $x \in A$ and equals $0$ if $x \not\in A$.

Using this we can write down densities of $\xi$ and $\eta$:

$\rho_{\xi}(x)=\mathbb{1}_{0\leq x \leq 1}(x)$,

$\rho_{\eta}(y) = \mathbb{1}_{y \geq 0}(y)\lambda e^{-\lambda y}$.

I assume $\mathbb{1}_{0\leq u \leq 1} := \mathbb{1}_{0\leq u \leq 1}(u), \mathbb{1}_{\frac{v}{u}\geq0} : = \mathbb{1}_{\frac{v}{u}\geq0}(v, u)$.

$\rho_{\xi\eta}(v) \stackrel{(1)}{=} \int_{-\infty}^{+\infty}\rho_{\xi}(u)\rho_{\eta}(\frac{v}{u})\frac{1}{|u|}du =\\ \int_{-\infty}^{+\infty}\mathbb{1}_{0\leq u \leq 1} \cdot \mathbb{1}_{\frac{v}{u}\geq0}\lambda e^{-\lambda v u^{-1}}|u^{-1}|du= \\ \int_{0}^{1}\mathbb{1}_{\frac{v}{u}\geq0}\lambda e^{-\lambda v u^{-1}}u^{-1}du = \mathbb{1}_{\{v\geq0\}}\int_{0}^{1}\lambda e^{-\lambda v u^{-1}}u^{-1}du$.

Equality (1) is just using formula of convolution of product.

If I used some uncommon notations or I can make my solution clearer, please let me know.

Answer 1

Too long for a comment.

Not sure why you integrate from $-\infty$ to $+\infty$ and there might be other problems.

This is how I would do it: \begin{align} &\mathbb P\big\{\xi\,\eta\le x\big\}=\textstyle\int_0^1\int_0^{x/u}\lambda\, e^{-\lambda y} \,dy\,du=\int_0^1 1-e^{-\lambda x/u }\,du\\[2mm] &=\textstyle 1-e^{-\lambda x}+\lambda x\,\Gamma(0,\lambda x) \end{align} (according to Wolfram Alpha) where $\Gamma(s,x)$ is the incomplete gamma function: $$ \Gamma(s,x)=\textstyle\int_x^\infty t^{s-1}e^{-t}\,dt\,. $$ Now differentiate that w.r.t. $x$ to get the PDF of $\xi\,\eta\,:$ $$ p_{\xi\eta}(x)=\textstyle\int_0^1 \frac{\lambda}{u} e^{-\lambda x/u}\,du=\lambda\,\Gamma(0,\lambda x)\,. $$

Answer 2

Here is another solution. Let $Z=\xi \eta$ with $\xi\sim\operatorname{Uniform}(0,1)$ and $\eta\sim\operatorname{Exponential}(\lambda)$. Then we have for the conditioned random variable $$ Z|\eta\sim\operatorname{Uniform}(0,\eta) $$ The joint distribution for $(Z,\eta)$ is $f_{Z,\eta}(z,\eta)=f_{Z|\eta}(z|\eta)f_\eta(\eta)$. Integrating the joint density w.r.t. $\eta$ should then give the distribution for $Z$. We write $$ \begin{aligned} f_Z(z) &=\int_0^\infty f_{Z|\eta}(z|\eta)f_\eta(\eta)\,\mathrm d\eta\\ &=\int_0^\infty \frac{1}{\eta}\mathbf 1_{z\in(0,\eta)} \lambda e^{-\lambda\eta}\,\mathrm d\eta\\ &=\int_0^\infty \frac{1}{\eta}\mathbf 1_{\eta\in(z,\infty)} \lambda e^{-\lambda\eta}\,\mathrm d\eta\\ &=\int_z^\infty \frac{1}{\eta} \lambda e^{-\lambda\eta}\,\mathrm d\eta\\ &=\lambda\int_{\lambda z}^\infty u^{0-1} e^{-u}\,\mathrm du\\ &=\lambda\Gamma(0,\lambda z), \end{aligned} $$ where $\Gamma(a,z)=\int_z^\infty t^{a-1}e^{-t}\,\mathrm dt$ is the (upper) incomplete gamma function. The is the same result obtained by @Kurt G and is the final solution to the product distribution. Note that this result does not simplify to elementary functions but you can express it in terms of $$ f_Z(z)=-\lambda\operatorname{Ei}(-\lambda z), $$ where $\operatorname{Ei}(x)=-\int_{-x}^\infty e^{-t}/t\,\mathrm dt$ is the exponential integral.