# $X,Y \sim \exp(1)$ and $U \sim [0,1]$ are independent, then what is $U(X+Y)$?

The Question:

Given random variables $X,Y \sim \exp (1)$ and $U \sim U[0,1]$ where $X$, $Y$, $U$ are independent, find the distribution of $U(X+Y)$.

My Attempt:

I have found that the density of $X+Y$ is

$$f_{X+Y}(v) = ve^{-v} \qquad v \in [0, \infty)$$

which I believe is correct.

However, when I try go on to finding the density $U(X+Y)$, I get

\begin{align} f_{U(X+Y)}(v) & = \int_v^\infty f_U \Big(\frac {v}{u} \Big) \cdot f_{X+Y}(u) \, du \\ & = \int_v^\infty (1) \cdot \big(ue^{-u} \big) \, du \\ & = \big[-ue^{-u}-e^{-u} \big]_v^\infty \\ & = (1+v)e^{-v} \end{align}

which isn't even a probability density function (it integrates to $2$).

What on earth am I doing wrong?

• @Clarinetist indeed you are correct – Henry May 22 '18 at 11:45

Since $X,Y\overset{iid}\sim\mathcal{Exp}(1)$ therefore $X+Y\sim\mathcal{Erlang}(2,1)$.   So, indeed, we have your result: $$f_{X+Y}(v)=ve^{-v}\mathbf 1_{v\geqslant 0}$$

However, since $U\sim\mathcal U[0;1]$, the product distribution should be $$\begin{split}f_{U(X+Y)}(s)&=\int_\Bbb R \tfrac 1 {\lvert t\rvert} f_U(\tfrac st)f_{X+Y}(t)\mathsf d t\\ &= \mathbf 1_{s\geqslant 0}\cdot\int_s^\infty \tfrac 1t \cdot 1\cdot te^{-t}\mathsf d t~\\ &= e^{-s}\mathbf 1_{s\geqslant 0}\end{split}$$

Which, of course, integrates to $1$.

Using the Jacobian change of variables theory: $$f_{UZ,Z}(s,t)=\begin{Vmatrix}\dfrac{\partial (s/t,t)}{\partial (s,t)}\end{Vmatrix}f_{U,Z}(s/t,t)=\dfrac{1}{\lvert t\rvert}f_{U,Z}(s/t,t)$$

• May I ask how did you get the $\dfrac{1}{|t|}$? – glowstonetrees May 22 '18 at 12:11
• Jacobian Determinant for the change of variables. – Graham Kemp May 22 '18 at 12:17

Let $M_X$, $M_Y$ be the moment-generating functions of $X$ and $Y$, respectively.

Then $$M_{X+Y}(t) = M_X(t)M_Y(t)=\left(\dfrac{1}{1-t}\right)^2$$ so $X+Y \sim \text{Gamma}(2, 1)$ - i.e., $$f_{X+Y}(v) = \dfrac{1}{\Gamma(2)1^{2}}v^{2-1}e^{-v/1}=ve^{-v}\text{, } v > 0\text{.}$$ Write $V = X + Y$ for now. We wish to find the density of $W = UV$. According to this, we have $$f_{W}(w) = \int_{-\infty}^{\infty}f_{V}(v)f_{U}\left(\dfrac{w}{v} \right) \cdot \dfrac{1}{|v|}\text{ d}v\text{.}$$ Observe that $$f_{U}(u) = \begin{cases} 1, & u \in [0, 1] \\ 0, & u \notin [0, 1] \end{cases}$$ so hence $$f_{U}\left(\dfrac{w}{v}\right)=\begin{cases} 1, & 0 < w/v < 1 \Longleftrightarrow 0 < w < v \\ 0, & w/v \leq 0 \text{ or }w/v \geq 1 \Longleftrightarrow w \leq 0\text{ or } w\geq v\text{.} \end{cases}$$ hence (note that the variable of integration is $v$) $$f_{W}(w)=\int_{w}^{\infty}ve^{-v}\cdot 1 \cdot \dfrac{1}{v}\text{ d}v$$ and note that $\dfrac{1}{|v|}$ turns into $\dfrac{1}{v}$ because we have $v > 0$. At last, $$f_{W}(w)=\int_{w}^{\infty}ve^{-v}\cdot 1 \cdot \dfrac{1}{v}\text{ d}v =\int_{w}^{\infty}e^{-v}\text{ d}v = e^{-w}$$ so $W \sim \text{Exp}(1)$.

• How did you get the extra $\dfrac 1v$? – glowstonetrees May 22 '18 at 12:14
• @glowstonetrees Click the link in my post. – Clarinetist May 22 '18 at 12:14