Question

According to the book to get the density function from a distribution function i take the derivate of F.
f(x,y) = d^2/dxdy F(x,y).
My understanding is that d^2/dxdy means that you first take the derivate with respect to x and then to y. Is my understanding correct? I got the right answer for (a) but the second one was completely off.
(a)=1/5
(b)=27/14sqrt(7)-20
For each of the following functions F, determine k such that F is a distribution function. Assume that all the probability lies in the indicated region.

a.F(x,y)=kxy(2x+3y) 0<=x<=1, 0<=y<=1
b.F(x,y)=kxy(2x+3y) x+y<=1, x>0, y>0

Answer 1

Usually \[\frac{\partial^2}{\partial x\partial y}F(x,y)=\frac{\partial}{\partial x}\left( \frac{\partial}{\partial y}F(x,y)\right) \], so you compute the derivative with respect to y first, then with respect to x. Although, in most cases, doing it with x first, then y gives the same result when you have nice "nice" functions, when second partial derivatives are continuous everywhere

Answer 2

Thanks for clearing that up and yes I do get the same derivative, k(4x+6y). \[\int\limits_{0}^{1}\int\limits_{0}^{1-x}k(4x+6y) dydx\] When i integrate this I get k=3/5 which is wrong according to the book. Any thoughts on that?

Answer 3

is it \(k=3/29\)

Answer 4

hm no wait that can't be right lol

Answer 5

hm yeah actually I get 3/5 as well.

Answer 6

What did the book say

Answer 7

book says \[k = \frac{ 27 }{14\sqrt{7}-20 }\]

Answer 8

wow how did they manage to get a square root in that

Answer 9

So it's either wrong or they're integrating at different points that are not \[\int\limits_{0}^{1}\int\limits_{0}^{1-x} f(xy) dydx\]

Answer 10

I get the same integral bounds though.

Answer 11

Maybe they are using the properties of the cdf to find this rather than going through a derivative route. ? Like:
\[\lim_{(x,y)\to(-\infty,-\infty)}F(x,y)=0\\
\lim_{(x,y)\to(+\infty,+\infty)}F(x,y) =1\\
\lim_{x\to -\infty}F(x,y)=0\\
\lim_{x\to +\infty}F(x,y)=F_Y(y)\\
\lim_{y\to -\infty}F(x,y)=0\\
\lim_{y\to +\infty}F(x,y)=F_X(x)\]
I'm not sure if that would give anything though... I can't really picture square roots coming out of that. :S ... Gosh I am not sure how else you could proceed

Answer 12

Thats alright thanks for the help. I will speak with my professor tomorrow about it and ill post what he tells me.

Answer 13

All right. Very curious about this problem now ;) (Hopefully it's a mistake hehe!)

Answer 14

Apparently the book is wrong and the answer is indeed k=3/5.

Answer 15

Oh glad to know we were right :)