Thread: Covariance and Correlation(continuous random variables)

Determine the value of c and the covariance and correlation for the joint probability density function f(x, y) = c for 0< x < 5, 0 < y and x - 1 < y < x + 1.

I am confused on how to set-up for the bounds for the integration. It looks like you have to split the region into 2 sections because of the 0 < y, but I am not sure. Thanks for any help.

Originally Posted by Oblivionwarrior

Determine the value of c and the covariance and correlation for the joint probability density function f(x, y) = c for 0< x < 5, 0 < y and x - 1 < y < x + 1.

I am confused on how to set-up for the bounds for the integration. It looks like you have to split the region into 2 sections because of the 0 < y, but I am not sure. Thanks for any help.

Can you draw the region on a graph. It's then very clear that the region needs to be split into two sections (which two sections will depend on what order of integration you choose. Personally I'd integrate first wrt y).

Either way you need two regions to integrate.
But to find c you just need the area and calculus is not needed for that.
You just find the area of the larger triangle and subtract off the area of the smaller one.

Ok. I found C to be 2/19. I integrated over the 2 regions. For the smaller region I got 3/2C, and for the larger region I got 8C. So 3/2C + 8C = 1, therefore C = 1/9.5 or 2/19.

To find the covariance would I use the formula E(XY) - E(X)E(Y)? To find E(XY) would I integrate again over the regions but this time have xy inside the integral? Would the C be 2/19 for both sides or would I need to split it up for both sides?

Originally Posted by Oblivionwarrior

Ok. I found C to be 2/19. I integrated over the 2 regions. For the smaller region I got 3/2C, and for the larger region I got 8C. So 3/2C + 8C = 1, therefore C = 1/9.5 or 2/19.

[snip]

As has already been suggested, since the joint pdf is constant you can use simple geometry:

You require c(area of trapezium - area of triangle) = 1.
Area of trapezium = 5(1 + 6)/2 = 35/2.
Area of triangle = (4)(4)/2 = 8.
c(35/2 - 8) = 1 => c = 2/19 so yes you're correct.

Originally Posted by Oblivionwarrior

[snip]
To find the covariance would I use the formula E(XY) - E(X)E(Y)? To find E(XY) would I integrate again over the regions but this time have xy inside the integral? Would the C be 2/19 for both sides or would I need to split it up for both sides?

Yes. Yes. Yes it would - c = 2/19 is the pdf so why on Earth would you split it up??