Simple marginal distribution problem

[FunkyDwarf]

Hey guys,

Doin revision for my maths exam and i came across this question from a past exam:

Homework Statement

Find fx(x,y) of
f(x,y) = \frac{(1+4xy)}{2} for 0 \leq x, y \leq 1 and zero otherwise

Homework Equations

Now this should equal\int \frac{(1+4xy)}{2} dyover all y but that leads to infinities ( as y goes from minus infinity to 1)which obviously we can't have. I am sure I am missing something simple and stupid i just need someone to point ito out :)

Cheers
-G

NOTE: Sorry this latex is stuffing up, tryin to fix it

 

[NateTG]

You might want to split the integral into a couple of pieces. Remeber:
f(x,y) \neq \frac{1+4xy}{2}

 

[ZioX]

I think what Nate is trying to say is that f(x,y) is only defined on a certain domain. Since f(x,y) is a pdf, y cannot be arbitrarily negative as this would make f(x,y) negative. Remember, the integral of f(x,y) over the domain must be 1.

Try to make sense of your domains. Draw them. X and Y can sometimes be dependant on each other...which can make things complicated.

 

[Dick]

Doesn't this expression,

0 \leq x, y \leq 1

mean x and y are both in [0,1]? If y runs to minus infinity the question doesn't make much sense.

 

[NateTG]

I think what Nate is trying to say is that f(x,y) is only defined on a certain domain. Since f(x,y) is a pdf, y cannot be arbitrarily negative as this would make f(x,y) negative. Remember, the integral of f(x,y) over the domain must be 1.

Actually, f(x,y) is only non-zero on a certain domain, it's defined on the entire plane.

 

[FrogPad]

A good thing to do, is first draw your "support". Sketch where the function is non-zero. This allows you to easily setup the bounds on the integral.

 

[ZioX]

Actually, f(x,y) is only non-zero on a certain domain, it's defined on the entire plane.

Oh come on! Grant me some liberties. Although you're right, I probably shouldn't have used those words.

 

[FunkyDwarf]

Ah of course! i did draw it, i just drew it wrong :P thanks guys