# Expected value of XY

I'm having trouble figuring out the expected value of xy.

Random variables x and y are described by the joint PDF:

$$f(x,y) = \begin{cases} K & \text{: if } x + y ≤ 1 , x > 0 , y > 0 \\ 0 & \text{: otherwise}\end{cases}$$

Determine the expected value of a random variable $r$ defined as $r=x y$ given that $\max(x,y) \leq 0.5$

I've tried to do: $\displaystyle \mathbb{E}(XY) = \iint_{\Omega} xy\cdot f(x,y) \,\mathrm{d}x\,\mathrm{d}y$, but I don't think it's correct.

The integral of f over the region where it is non-zero is K * the area of the right triangle

with sides both of length 1 and so is K * 1/2. This must be 1, so K = 2.

The expected value of X Y given that X and Y are both <= 1/2 is he double integral of x y * 2

over the the square where x, y are <= 1/2. This is 2 * x^2 / 2 * y^2 / 2 for x = y = 1/2.

So the final answer is 1/32.

Note that $(X,Y)$ is uniformly distributed on the triangle $$T=\{(x,y)\mid x\gt0,y\gt0,x+y\lt1\}$$ and that the domain $$S=\{(x,y)\mid x\gt0,y\gt0,\max(x,y)\lt\tfrac12\}$$ is the square $(0,\frac12)^2$, which is fully included in $T$. Thus, $(X,Y)$ conditioned on $\max(X,Y)\lt\frac12$ is uniformly distributed on $S$. As such, $$E(XY\mid\max(X,Y)\lt\tfrac12)=E(UV),$$ where $U$ and $V$ are i.i.d. uniform on $(0,\frac12)$, hence $$E(XY\mid\max(X,Y)\lt\tfrac12)=E(U)^2=\left(\tfrac14\right)^2=\tfrac1{16}.$$

• Doh! Right. We can't use the double integral of $xy\cdot f(x,y)$ because we need to use the conditional distribution ($f(x,y\mid x\leq \frac 12, y\leq \frac 12)$ such that $\iint_\Omega f(x,y\mid x\leq \frac 12, y\leq \frac 12) \mathrm{d}x\mathrm{d}y = 1$). – Graham Kemp Apr 25 '14 at 9:32