0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$ E(XY\mid\max(X,Y)\lt\tfrac12)=E(U)^2=\left(\tfrac14\right)^2=\tfrac1{16}. $$

$\endgroup$
  • $\begingroup$ 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 $\iint_\Omega f(x,y\mid x\leq \frac 12, y\leq \frac 12) \mathrm{d}x\mathrm{d}y = 1$). $\endgroup$ – Graham Kemp Apr 25 '14 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.