$$f(x,y) = \frac{x+y}{2}, \text{where } x>0,\ y>0,\ 3x + y<3 $$

$$ \text{Find the CDF of } V=XY,\ F_{V}(v) $$

I have found the that $V$ is bounded from $0$ to $\frac34$, but don't understand how to solve for the limits of the integral. I am currently attempting to solve for the intersection between $V=XY$ and the support of the pdf. Is there a more convenient way to solve for the CDF of $V$? What is the best way to think about this kind of problem?

  • $\begingroup$ I have not check your calculation, but it is correct that you check the support of the resulting random variable $V$ first. Then you know the values of the CDF outside the support. When you consider the value within the support, in this question you are given the joint, so the most direct way is what you have did - to compute $\Pr\{XY < v\}$ using the given joint. If you visualize it on the $x-y$ plane, then you are finding the area of the intersection of the region under the hyperbola and the triangular support. $\endgroup$
    – BGM
    Sep 28, 2017 at 5:45

1 Answer 1


The region where $f$ is defined is bounded by a triangle $\Delta$ between the two axes and the line L: $3x+y=3$. We can check that $\int \int_\Delta f dxdy =1$. The level sets of $V$ are hyperbolae. For values $V< \frac{3}{4}$, the hyperbola intersects the line L at two points, call them $x_1$ and $x_2$. Solving the equations of the line and the hyperbola together for a given $V=v$, $$x_1=\frac{1-\sqrt{1-\frac{4v}{3}}}{2}; x_2=\frac{1+\sqrt{1-\frac{4v}{3}}}{2};$$ enter image description here $F_V(v)$ is the area within the triangle, to the left of the hyperbola $V=v$ (shown in black). Which is same as $1-A$ where A the area to the right of the hyperbola but below the line L (shown in blue).

$$F_V(v) = 1- \int_{x_1}^{x_2} \int_{\frac{v}{x}}^{3-3x} \left(\frac{x+y}{2}\right) dy dx $$ and you're done.

$$F_V(v) = 1 -\frac{1}{2}\left[ \frac{x^3}{2}-3x^2+(\frac{9}{2}-v)x+\frac{v^2}{x} \right]_{x_1}^{x_2}$$

You should able to take it from here (the above expression now contains only $v$). It's straightforward from here, though a bit ugly when I worked out each of the terms. Tip: making the substitution $a \equiv \sqrt{1-\frac{4v}{3}}$ makes the subsequent algebra quite a bit less ugly.


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