joint distribution of f(x,y)

[hphsu]

joint distribution

f(x,y) = (1/4)xy if 0<x<2, 0<y<2
= 0 otherwise

how to find P(XY<2)?

I wrote it out as ∫(from 2 to 0) ∫(from (2/x) to 0) (1/4)xy dy dx,
but my answers never worked out.
I think the problem lies in integrating from what to what, but I'm not sure how to correct the mistake. Please help! Thanks!

[Dason]

If x < 1 you would be integrating y from 0 to ... something larger than 2. But you have additional constraints that tell you that y <= 2. You need to account for that.

[hphsu]

If x < 1 you would be integrating y from 0 to ... something larger than 2. But you have additional constraints that tell you that y <= 2. You need to account for that.

So now I thought of P(XY>2) = 1 - P(XY<= 2), which would account for that problem.

I calculated with 1 - (what I wrote out on the first thread), which still doesn't work. Can someone please tell me why? Thanks.

[Dason]

So now I thought of P(XY>2) = 1 - P(XY<= 2), which would account for that problem.

I calculated with 1 - (what I wrote out on the first thread), which still doesn't work. Can someone please tell me why? Thanks.

I don't get what you're saying. But I suggest just drawing out the support of the distribution and then drawing the area you want to integrate. Then break that into easily integrable chunks.

[hphsu]

alright I'll try to do that again, thanks!