# How to find CDF of a function of continuous joint distribution from PDF of joint distribution?

Here's what I think I should proceed:

1. Make a level curve for the function keeping the constraints given in the PDF of joint distribution in mind.

2. Find the area of interest keeping the constraints given in the PDF of joint distribution in mind.

3. Evaluate the integral using $$w$$ in the limits.

I also think the following should be correct:

1. Given a function in $$x$$ and $$y$$,

$$F_W(w)=P[W\le w]=P[Y\le \text{ express }y\text{ in terms of }w\text{ and } x]=F_Y(y)$$

Now substitute $$y$$ in terms of $$w$$ and $$x$$.

So we need to first calculate $$F_Y(y)$$ and then simply plug in $$y$$ as expressed in terms of $$w$$ and $$x$$. Is this correct?

If $$(X,Y)$$ has the pdf $$f$$ and $$g$$ is any (measurable) function of $$X$$ and $$Y$$, then by definition CDF of $$g$$ is

\begin{align} P(g(X,Y)\le z)&=E\left[\mathbf1_{g(X,Y)\le z}\right] \\&=\iint \mathbf1_{g(x,y)\le z}f(x,y)\,\mathrm{d}x\,\mathrm{d}y\qquad\,,\small\text{ for all }z\in\mathbb R \end{align}

This is the same as saying $$P(g(X,Y)\le z)=P((X,Y)\in A)$$ where $$\small A=\{(x,y):g(x,y)\le z\}$$.

Now of course the evaluation of the double integral would depend on the specific structure of $$f$$ and $$g$$. And as is often true for double integrals, making a sketch of the domain of integration might help you determine the limits of integration easily. You might even need a change of variables, but that is part of the general process of evaluating multiple integrals.

To see what type of calculations this could involve, take the answers in this post for example where we find the CDF of $$\sqrt{X^2+Y^2}$$ when $$(X,Y)$$ has the pdf $$f(x,y)=\mathbf1_{0.

• What do you mean by Support? Also is my second approach correct, ie. Express Y in terms of w and X and then simply evaluate using the CDF of Y? – Adarsh Kumar Mar 24 at 9:37
• I don't understand your approach. By 'support', I mean this. – StubbornAtom Mar 24 at 9:49