# Finding Covariance of simple joint DF

• Sep 12th 2012, 12:32 AM
Finding Covariance of simple joint DF
A joint density function is given by:
f(x, y) = kx 0 < x, y < 1
0 otherwise
Find Cov(X, Y)
******
Those bounds are giving me trouble. At first I read it as 0 < x < y < 1. But I get 2 different values for k depending if I use the direction of f(x) or f(y)...

I found:
f(x) = k - kx^2 which lead to k = 3/2
f(y) = (ky^2)/2 which lead to k = 6
I'm not sure if it's possible to have 2 different values of k like that ?
E(xy) led to another problem, which k to chose ? So I must have done something wrong with those bounds...

Thanks !!!
• Sep 12th 2012, 07:38 AM
harish21
Re: Finding Covariance of simple joint DF
Quote:

A joint density function is given by:
f(x, y) = kx 0 < x, y < 1
0 otherwise
Find Cov(X, Y)
******
Those bounds are giving me trouble. At first I read it as 0 < x < y < 1. But I get 2 different values for k depending if I use the direction of f(x) or f(y)...

I found:
f(x) = k - kx^2 which lead to k = 3/2
f(y) = (ky^2)/2 which lead to k = 6
I'm not sure if it's possible to have 2 different values of k like that ?
E(xy) led to another problem, which k to chose ? So I must have done something wrong with those bounds...

Thanks !!!

No you cant have 2 different values of k.

you are given that $f(x,y) = k\cdot x\;\;0

that means $0; x and y range between 0 and 1, but you are not given that x<y.

so $f(x) = \int_y f(x,y) dy= \int_0^1 kx \; dy= kx$

$f(y)=\int_x f(x,y) dx = \int_0^1 kx dx = \frac{k}{2}$

now you can find the value of k by using the fact that $\int_0^1 f(x)\;dx = 1$ and $\int_0^1 f(y) dy =1$

you can then plug the value of k in f(x,y) and integrate to see that it integrates to 1.
• Sep 12th 2012, 10:50 AM