Thread: 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 !!!
The Answer is 0

Originally Posted by TheCountador

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 !!!
The Answer is 0

No you cant have 2 different values of k.

you are given that

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

so



now you can find the value of k by using the fact that and

you can then plug the value of k in f(x,y) and integrate to see that it integrates to 1.

I've got it thank you. That comma in the range threw me off

but k cannot be 0.

edit: Ignore this post. looks like your covariance is 0. I thought you were saying k=0