Show that $X+Y$ and $X/Y$ are independent.

Question

Here's my problem:

Let $X$ and $Y$ be iid with a common pdf

$f(t) = exp(-t),$ if $t > 0$,

$0$ otherwise.

Show that $X+Y$ and $X/Y$ are independent.

Here's what I'm thinking:

$X+Y$ and $X/Y$ are independent if $M(t_1,t_2) = E[e^{t_1(X+Y) + t_2(X/Y)}] = E[e^{t_1X}]E[e^{t_1Y}]E[e^{t_2(X/Y)}]$, i.e. the joint mgf of $X+Y$ and $X/Y$ is equal to the product of the individual mgfs $M_{X+Y}(t)$ and $M_{X/Y}(t)$.

We can easily show that the mgf $M_{X+Y}(t) = M_X(t)M_Y(t) = E[e^{tX}]E[e^{tY}] = e^{-(x+y)} (x>0,y>0)$, which leaves me to solve for $M_{X/Y}(t) = E[e^{t_2(X/Y)}]$. But since $X/Y$ is a fraction, it remains difficult to solve.

Am I doing this the right way? Is there a better way? If not, how do I complete this?

Thanks

Answer 1

You are correct, the variables are independent if you can show that $$ E[e^{t_1 (X+Y) + t_2 (X/Y)}]= E[e^{t_1 (X+Y)}] E[e^{t_2 (X/Y)}] . $$ This does not hold unconditionally and its validity relies on the hypothesis at hand. A.S. gave a beautiful proof in the comment. Alternatively you can reason as you where doing:

First compute

\begin{align} E[e^{t_1 (X+Y) + t_2 (X/Y)}]&= \int_0^\infty \!\! \!dx \int_0^\infty \!\! \! dy e^{-x(1-t_1 -t_2/y) - y(1-t_2)} \\ &= \int_0^\infty \!\! \! dy \frac{e^{-y(1-t_1)}}{1-t_1 -t_2/y} \end{align}

Now use the substitution $y(1-t_1)=z$ to obtain

$$ M(t_1,t_2) = \frac{1}{(1-t_1)^2}\int_0^\infty \!\! \!dz \frac{e^{-z}}{1-t_2/z} $$

Now you can check that \begin{align} E[e^{t_1 (X+Y)}] E[e^{t_2 (X/Y)}] &= \int_0^\infty \!\! \!dx \int_0^\infty \!\! \! dy e^{-x(1-t_1) -y(1-t_2)} \int_0^\infty \!\! \!dx \int_0^\infty \!\! \! dy e^{-x(1-t_2/y)-y} \\ &= \frac{1}{(1-t_1)^2}\int_0^\infty \!\! \!dy \frac{e^{-y}}{1-t_2/y} \end{align}

and you are done.

By the way, note that the MGF of your exponential distribution is $1/(1-t_1)$ (for $t_1 <1$).

Answer 2

Let $U=X+Y$ and $V=\frac XY$. It is clear that $U\sim\operatorname{Exp}(2)$. If $v>0$ then $vY\sim\operatorname{Exp}\left(\frac1v\right)$, so $$\mathbb P(V \leqslant v) = \mathbb P(X\leqslant vY) = \frac1{1+\frac1v} = \frac v{1+v}.$$ Therefore $U=g(X)$ and $V=h(Y)$ where \begin{align} g(x) &= 2x\\ h(y) &= 1+e^{\frac12 y}, \end{align} and hence $\sigma(U)\subset\sigma(X)$, $\sigma(V)\subset\sigma(Y)$. Since $X$ and $Y$ were assumed to be independent, it follows that $U$ and $V$ are independent.