Joint Density Formulas for XY and X/Y

Question

I used convolution to find a formula for $X+Y$, but am unsure on how to figure out formulas for $XY$ and $X/Y$. Any help would be appreciated.

Answer 1

Convolution only works for independent X,Y. In general the formulas will involve the joint density, e.g. for X+Y: $$ f_{X+Y}(s) = \int_\mathbb R f_{X,Y}(s-y,y)dy $$ and so on... Note that when X,Y are independent, we have $f_{X,Y}(s-y,y) = f_X(s-y)f_Y(y)$ and we recover the convolution integral $f_X * f_Y$

Answer 2

I'll solve a more general question for you here and leave it to you to tailor it to your needs. Note that your solution will not be nearly as complex as what I'll present below. The reason for this complexity is to make sure nothing can go wrong. Here are the creatures appearing in this solution:

1. Let $G:D\to \mathbb{R}$ be a continuously differentiable function, with $D\subset \mathbb{R}^2$ some subset. For example for $G(x,y)=x/y$ we take $D=\{(x,y)\mid y\neq 0\}$.
2. Let $X,Y$ two continuous random variables with joint probability density $\rho(x,y)$.

We are interested in finding the probability density of random variable $G(X, Y)$ when the values of $(X,Y)$ lie in the domain of $G$, i.e. $D$.

3. The values of the random variable $G(X,Y)$ will be denoted by $\zeta$. The probability density will be denoted by $\rho_G(\zeta)$.

We will impose some extra conditions/notations to make life simpler:

• Let $U\subset \mathbb{R}$ be an open subset such that $\partial G(x,y)/\partial y\neq 0$ for all $(x,y)\in T$, where $T:=(U\times \mathbb{R})\cap D$.
• We denote by $Z\subset \mathbb{R}^2$, the set of values $\zeta$ such that if $(x,y)\in D$ is such that $G(x,y)=\zeta$ then $(x,y)\in T$. We also assume that complement of $Z$ in the space of values of $G(X,Y)$ is measure zero (i.e. we do not have to worry about a $\zeta\notin Z$).

The first condition is a version of implicit function theorem. The consequences of the above are:

• Inside $T$ one can solve $G(x,y)=\zeta$ by $y=h_\zeta(x)$ with $h_\zeta:U\to \mathbb{R}$ continuous.
• If $\zeta\in Z$ and $G(x,y)=\zeta$ and $G(x,y_1)=G(x,y_2)=\zeta$, then $y_1=y_2=h_\zeta(x)$.

For your question, $G(x,y)=x+y,\:xy, \:x/y$ respectively. I'll let you figure out what $U,D,T$ and $Z$ are respectively.

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OK, now let's solve the question. The probability of $G(X,Y)\in [\zeta, \zeta+\delta \zeta]$ is equal to $\rho_G(\zeta)\delta\zeta$ by definition (we are taking $\delta \zeta$ as an infinitesimal). Let $R_\zeta\subset \mathbb{R}^2$ be the set of points $(x,y)\in \mathbb{R}^2$ that satisfy $G(x,y)\in [\zeta, \zeta+\delta \zeta]$. Then $$ \rho_G(\zeta)\delta\zeta = \int_{R_\zeta} \rho(x,y)\:dx\:dy $$ Assuming $\zeta\in Z$, we have $R_\zeta\subset U$. Change of variables $(u,v)=(x,G(x,y))$ yields $$ du\: dv = |\begin{vmatrix} 1 & 0\\ \partial G/\partial x & \partial G/\partial y \end{vmatrix}| = |\partial G/\partial y|\: dx\: dy\Longrightarrow dx \: dy = \frac{du \: dv}{|\partial G/\partial y|} $$ Note that $R_\zeta$ in terms of our new variables is just $\zeta\leq v\leq \zeta+\delta \zeta$ (and $u=x$ completely free in $U$). Therefore $$ \rho_G(\zeta)\delta \zeta = \delta \zeta \int_U \frac{\rho(x,h_\zeta(x))dx}{|\frac{\partial G}{\partial y}(x,h_\zeta(x))|} \Longrightarrow \boxed{ \rho_G(\zeta) = \int_U \frac{\rho(x,h_\zeta(x))dx}{|\frac{\partial G}{\partial y}(x,h_\zeta(x))|} } $$

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Alternatively, if you are familiar with delta "function", one has the following: $$ \boxed{ \rho_G(\zeta)=\mathbb{E}_{(X,Y)}[\delta(\zeta - G(X,Y))]} $$ which works in even a more general setting than above (that is if you know how to deal with delta "functions" correctly).