# compute probability density function of a bivariate function without sampling

Suppose $X_1 \sim f_{X_1}(x_1)$, $X_2 \sim f_{X_2}(x_2)$ are random variables with known probability density function.

Is there any way to compute the probability density function of a bivariate function $g(x_1,x_2)$, assuming on specific finite support $x_1 \in [a_1,b_1]$ and $x_2 \in [a_2,b_2]$ without Monte Carlo sampling or any other sampling method? $X_1$ and $X_2$ are independent, and $g$ is "smooth".

I'm looking for a way to possibly implement the computation numerically for an arbitrary $g$.

• what do you know of the relationship between $X_1,X_2$? E.g. if they are independent, $g(x_1, x_2) = f_{X_1}(x_1) \times f_{X_2}(x_2)$... – gt6989b Apr 21 '16 at 17:34
• yes, assuming they are independent. thanks for pointing it out! – kensaii Apr 21 '16 at 17:34
• why not condition on $X_2$? – gt6989b Apr 21 '16 at 17:40
• I'm not so familiar with Bayesian rules conditioning... can you please elaborate? :) – kensaii Apr 21 '16 at 17:42

Let's rename your variables $X,Y$ for easier notation. So you are defining $$Z = g(X,Y)$$ and asking what is the probability density function of $Z$. For simplicity, let's assume a particular (very simple) $g$, let's say $Z = X+Y$.
$Z=z$ means $X+Y=z$, so $X=z-Y$; let's condition on $Y=y$, then you end up with $$f_Z(z) = \int_\mathbb{R} f_X(z-y) f_Y(y) dy.$$
So if $g(x,y)$ has a nice form allowing to solve $z = g(x,y)$ for one of the variables in terms of the other, you can use this technique to get the pdf...
• thank you! this is also known as the convolution formula, right? what if $g$ is not a one-to-one transformation? In this case the Jacobian is 1 so it becomes quite straightforward... – kensaii Apr 22 '16 at 13:50