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If $Y$ is a random variable defined as $Y=g(X_1,X_2)$, where $X_1$ and $X_2$ are two different random variables whose distributions are known (say with pdf's $f_{X_1}$ and $f_{X_2}$), how do we find the distribution of $Y$ (i.e. the pdf $f_Y$)?

Is there a general method to solve this problem? If there isn't, let us consider two specific cases which I am interested in.

a) $Y=X_1X_2$; $X_1$ and $X_2$ being independent random variables.

b) $Y=X_1X_2$; $X_1$ and $X_2$ being correlated random variables.

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closed as off-topic by Did, Travis, colormegone, dustin, graydad Jan 22 '15 at 4:18

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The general method for solving is just to write it down and compute:

$$F_Y(y)=P(Y\le y)=P(g(X_1,X_2) \le y) = \int_{g(u,v) \le y} f_{X_1,X_2}(u,v)\; dv \, du$$

If $g(u,v)=u v$ (and assuming only positive values) then

$$F_Y(y) = \int_0^{\infty} \int_0^{y/u} f_{X_1,X_2}(u,v)\; dv du$$

If they are independent

$$F_Y(y) = \int_0^{\infty} f_{X_1}(u) \int_0^{y/u} f_{X_2}(v)\; dv \, du =\int_0^{\infty} f_{X_1}(u) \, F_{X_2}(y/u) \, du $$ and

$$f_Y(y) = \int_0^{\infty} \frac{1}{u}f_{X_1}(u) \, f_{X_2}(y/u) \, du $$

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