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I ran into an old exercise but I seem to have messed up somehow. Can you tell me what went wrong?

Let $U \sim \mathrm{Unif}(0,1)$ and $V \sim \Gamma(2,1)$ with $U,V$ independent. Show that $UV$ has the distribution $\mathrm{Exp}(1)$.

I attempted this with a substitution $S = UV$ and $T = V$. I know that $U,V$ have joint probability density function $f_{U,V}(u,v) = ve^{-v} $. Blindly using change of variables, I obtain the Jacobian $|J| = \frac{1}{t} $ and so $f_{S,T}(s,t) = e^{-t} $. Which seemingly implies $S = UV$ has uniform distribution and $T$ has exponential distribution $\mathrm{Exp}(1)$. But this makes no sense because $T = V$ and $T$ has distribution $\Gamma(2,1)$.

My notes tell me the substitution $S = UV$ and $T = \frac{U}{V}$ give the right answer. But what is it that makes my substitution invalid and what went wrong with my working?

E.Lim
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3 Answers3

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Brute force: let $S = UV$. Then $$\begin{align*} \Pr[S \le s] &= \Pr[UV \le s] = \Pr[V \le s/U] = \int_{u=0}^1 \Pr[V \le s/u] f_U(u) \, du \\ &= \int_{u=0}^1 \int_{v = 0}^{s/u} \frac{b^a v^{a-1} e^{-bv}}{\Gamma(a)} \, dv \, du \\ &= \frac{b^a}{\Gamma(a)} \left( \int_{v=s}^{\infty} \int_{u=0}^{s/v} v^{a-1} e^{-bv} \, du \, dv + \int_{v=0}^s \int_{u=0}^1 v^{a-1} e^{-bv} \, du \, dv \right) \\ &= \frac{b^a}{\Gamma(a)} \left( \int_{v=s}^\infty s v^{a-2} e^{-bv} \, dv + \int_{v=0}^s v^{a-1} e^{-bv} \, dv \right). \end{align*}$$ For $a = 2$, $b = 1$, we get $$\Pr[S \le s] = s \int_{v=s}^\infty e^{-v} \, dv + \int_{v=0}^s v e^{-v} \, dv = s e^{-s} + (1 - (1+s)e^{-s}) = 1 - e^{-s},$$ hence $S$ is exponential with rate parameter $1$, as claimed.

heropup
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You have the right joint density function provided you get the right boundaries: $$ f_{S,T}(s,t) = \begin{cases} e^{-t} & \text{if }0<s<t, \\ 0 & \text{otherwise.} \end{cases} $$ (You can draw a picture and see that this is half of the first quadrant.)

Now suppose we want the marginal distributions of $S$ and $T$. $$ f_S(s) = \int_s^\infty f_{S,T}(s,t)\,dt = \int_s^\infty e^{-t}\,dt = e^{-s}\text { if }s>0. $$ So $S$ is exponentially distributed with expected value $1$. $$ f_T(t) = \int_0^t f_{S,T}(s,t)\,ds = \int_0^t e^{-t} \,ds = te^{-t}\text{ if }t>0. $$ So $T$ has the same Gamma distribution that $V$ has.

Perhaps some further insight can be afforded by two additional exercises:

  • Show that the distribution of $T-S$ is the same as the distribution of $S$.
  • Show that $T-S$ and $S$ are independent.

From these it follows that $T$ is the sum of two independent exponentially distributed random variables.

BTW, the phrase "product of distributions" isn't quite right. This is about the distribution of a product, not about the product of distributions.

Michael Hardy
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If $U\sim U(0,1)$ then $f_U(u)=1 I_{[0,1]}(u)$ and $V\sim\Gamma(2,1)$ then $f_V(v)=ve^{-v}I_{[0,\infty]}(v)$

Now take $W=g_1(U,V)=UV$ and $Z=g_2(U,V)=V$ thus $$J=\begin{bmatrix}\frac{\partial W}{\partial U}&&\frac{\partial W}{\partial V}\\\frac{\partial Z}{\partial U}&&\frac{\partial W}{\partial V}\end{bmatrix}=v\Rightarrow |J|^{-1}=\frac{1}{v}$$ You know that $$f_{W,Z}(w,z)=f_{U,V}(u,v)(g_1^{-1}(w,z),g_2^{-1}(w,z))|J|^{-1}=e^{-z}I_{[0,\infty]}(z)$$ $$f_Z(z)=\int_0^1e^{-z}dw=e^{-z}I_{[0,\infty]}(z)$$

Why integrate from $[0,1]$? Since that $U$ is defined from $[0,1]$ and $V$ in $[0,\infty]$ then the product $UV$ It is defined only between $[0,1]$ because if $U>1\Rightarrow UV=0$

I'm not completely sure of my idea since a while since I probability.

Roland
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