1. ## Continuous joint probabilities

Suppose x,y ~ f(x,y) = k(x-y) given 0=< X =< Y =< 1, otherwise it = 0
What value of k makes this a density?

I'm not quite sure what I'm being asked to find...I know it asks for k, but what constraints/definition is needing to be met in order for this to be a density? I just set up a double integral, each from 0 to 1, of k(x-y) but came out in the end with k*0, so I think this is somewhat incorrect...can anyone help me? I just need to know which direction to move in. Thanks!

2. Originally Posted by mistykz
Suppose x,y ~ f(x,y) = k(x-y) given 0=< X =< Y =< 1, otherwise it = 0
What value of k makes this a density?

I'm not quite sure what I'm being asked to find...I know it asks for k, but what constraints/definition is needing to be met in order for this to be a density? I just set up a double integral, each from 0 to 1, of k(x-y) but came out in the end with k*0, so I think this is somewhat incorrect...can anyone help me? I just need to know which direction to move in. Thanks!
You have to find the value of k such that $\displaystyle \int_{x = 0}^{1} \int_{y = x}^{y = 1} k (x - y) \, dy \, dx = 1$.

Do you see why?

3. Would that be that, because it's a density, it needs to integrate to 1 overall?
and since y is always greater than or equal to x, but less than 1, that gives the bounds

4. Originally Posted by mistykz
Would that be that, because it's a density, it needs to integrate to 1 overall?
and since y is always greater than or equal to x, but less than 1, that gives the bounds
Yes.

5. Alright, I understand it then. Thank you!

6. Originally Posted by mistykz
Suppose x,y ~ f(x,y) = k(x-y) given 0=< X =< Y =< 1, otherwise it = 0
What value of k makes this a density?

I'm not quite sure what I'm being asked to find...I know it asks for k, but what constraints/definition is needing to be met in order for this to be a density? I just set up a double integral, each from 0 to 1, of k(x-y) but came out in the end with k*0, so I think this is somewhat incorrect...can anyone help me? I just need to know which direction to move in. Thanks!

you just need to perform the double integral and set it equal to one.
$\displaystyle k\int_0^1\int_0^y (x-y) dxdy$.
I prefer integrating x first since that makes both lower bounds zero,
but you can integrate in either order.