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

Re: Finding Covariance of simple joint DF

Quote:

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 $\displaystyle f(x,y) = k\cdot x\;\;0<x,y<1$

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

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

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

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

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

Re: Finding Covariance of simple joint DF

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

Re: Finding Covariance of simple joint DF

but k cannot be 0.

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