Double integration

• Dec 12th 2012, 01:16 PM
zeubi94
Double integration
Hello, do you mind helping me with this please?

let for some a>0

f(x,y) = a2(cos (x - y) + sin ( x + y)) 0 ≤ x π/2 and 0 ≤ y π/2

Assume without computing a that f(x,y) is the p.d.f of a bivariate random variable ( X,Y) and compute the marginal p.d.f f1(x) and f2(y). Either from f1(x) or f,(y) find the value a > 0 such that f(x,y) is a proper p.d.f
So i need to find a such as the function equals to 1.
i should find 1/2 but i must have forgotten something..
• Dec 12th 2012, 01:40 PM
zeubi94
Re: Double integration
any ideas?
• Dec 12th 2012, 02:37 PM
Plato
Re: Double integration
Quote:

Originally Posted by zeubi94
let for some a>0
f(x,y) = a2(cos (x - y) + sin ( x + y)) 0 ≤ x π/2 and 0 ≤ y [COLOR=#000000][FONT=sans-serif]π/2

$\int_0^{\frac{\pi }{2}} {\int_0^{\frac{\pi }{2}} {\left[ {\cos \left( {x - y} \right) + \sin \left( {x + y} \right)} \right]dydx} } = 4$
• Dec 12th 2012, 02:43 PM
zeubi94
Re: Double integration
Thanks for your answer but do you mind explaining by steps because i am actually stuck somewhere?!
• Dec 12th 2012, 02:46 PM
Plato
Re: Double integration
Quote:

Originally Posted by zeubi94
Thanks for your answer but do you mind explaining by steps because i am actually stuck somewhere?!

Yes I do mind. But you can look at this page.
• Dec 12th 2012, 03:20 PM
zeubi94
Re: Double integration
Thanks but i should find 1/4 instead of 4 :s