@vv

Not really a severe limitation of **PDF**.

Or you have to consider in this case that "**int** has severe limitations" too ...

The problem here reduces to the integration of some function z=f(x,y) over a domain D(x,y | z) where "|" stands for "conditional to the value z of Z".

Generally difficult, without even considering the fact that the CDF of a sum of independent random variable already invokes a convolution product.

A few words about your toy-case to explain all of this.

Let **Z = X^2 + Y^2 + a*X*Y**.

First point: it is easy to show that the level curves Z = constant are ellipses as long as |a| < 2 (hyperbolas for |a| > 2 and a degenerate conocs for |a|=2).

Second point: there are a few obvious cases that makes obvious the computation of Prob(Z < z)

- a=2 (
**Z = (X+Y)^2**) here the level curves of Z are parallel lines of slope -1
- a=-2 (
**Z = (X-Y)^2**) here the level curves of Z are parallel lines of slope -1
- a=0 (
**Z = X^2 + Y^2**) here the level curves of Z are concentric circles of center (0,0)

Apart those 3 elementary situations (Maple 2015 doesn't even solve without a little bit of help), let us consider a simpler problem than your original one in order to introduce a geometrical approach.

Let X any be two Uniform random variable over [-H, H] where H is some strictly positive number.

Then, provided z is "sufficiently small" regarding to H, Prob(Z < z) is just the surface of some ellipsa.

Under the change of variables {X=U*cos(Pi/4)+V*sin(Pi/4), Y=-U*sin(Pi/4)+V*cos(Pi/4)} **Z = X^2 + Y^2 + a*X*Y**

writes** Z = (1-(1/2)*c)*U^2+(1+(1/2)*c)*V^2.**

With A = 1/sqrt((1-(1/2)*c)) and B = 1/ sqrt((1+(1/2)*c)) this relations becomes **Z = U^2/A^2+V^2/B^2**

Then Prob(Z < z) = Pi*(A*z)*(B*z).

This result holds only if ellipsa **U^2/A^2+V^2/B^2 = z^2 **is entirely contained in the unit square [-H, H]^2.

Now a more complex case.

First observe that **U^2/A^2+V^2/B^2 = z^2 **describe ellipses centered at point (0,0): then there exist no complete ellipsa within the square (X,Y)=[0, H]^2 and Prob(Z < z) is the measure of some angular sector of the ellipsa defined by angles alpha and beta.

A general formula for this measure, with a proper definition of alpha and beta, is

A*B/2*(arctan(A/B*tan(alpha)) - arctan(A/B*tan(beta))) (... = Prob(Z < z) )

Even if it is not very difficult to compute "by hand", you already see that some auxiliary difficulties occur:

- express
**Z = X^2 + Y^2 + a*X*Y **in the more suitable form **Z = (1-(1/2)*c)*U^2+(1+(1/2)*c)*V^2**
- identify the ellipsa
**U^2/A^2+V^2/B^2 = z^2 **
- compute alpha and beta

It is not very surprising that **PDF** (or **int**) is not capable to do tje job without any human help.

To give you a better idea of the details of the computations you can look to the attached file.

It doesn't contain a complete solution for any value of a in [-2, 2], but just an example for the case a=4/3 that **PDF** doesn't handle (generalisation of the given example is straightforward).

ToyProblem.mw

If you want a more general solution for any value of "a", you have to analyse the different situations when the iso-probability curves are hyperbolas.

At the end a complete solution of your toy-problem becomes a very complex issue.