# Finding covariance using the joint pdf

1. Jun 9, 2012

### Flashcop

The following question appeared on a practice exam:

For
f(x,y) = 24xy if 0<x+y<1 , 0<x,y
0 elsewhere

find Cov(X,Y)

I used Cov(X,Y) = E(XY) - E(X)E(Y) to calculate covariance, with

E(XY) = $\int^{1}_{0}\int^{1-y}_{0}24x^{2}y^{2}dxdy$

but for some reason I didn't get the suggested answer of 1/15

Can someone explain to me what I did wrong? Was there something wrong with the terminals?

2. Jun 9, 2012

### chiro

Hey Flashcop and welcome to the forums.

If I were you I would in later cases use the alternative definition COV(X,Y) = E[[X-E[X][Y-E[Y]] and calculate (X - E[X])(Y - E[Y])f(x,y)dxdy for the integral to double check your calculations if you think something is awry (it is redundant but it is a good way to cross-check your calculations in the case of say an algebraic mistake).

For this problem, I think the best way to get relevant advice is to go through your calculation step by step and post it here so that we can go through your algebra and your reasoning.

3. Jun 10, 2012

### Flashcop

Thanks mate :)

So using the alternate definition, I got:

Cov(X,Y) $= 24 \int^{1}_{0}\int^{1-y}_{0}(x-\frac{2}{5})(y-\frac{2}{5})xydxdy$
$= 24 \int^{1}_{0}\int^{1-y}_{0}x^{2}y^{2}-\frac{2x^{2}y}{5}-\frac{2xy^{2}}{5}+\frac{4xy}{25}dxdy$
$=24\int^{1}_{0}\frac{-1}{75}(2-5y)^{2}(y-1)^{2}ydy$
$=\frac{-24}{75}\int^{1}_{0}(2-5y)^{2}(y-1)^{2}ydy$
$=\frac{-24}{75}*\frac{1}{12} =\frac{-2}{75}$

Which is the same answer I got before, so that probably rules out any algebraic errors.