# Simple marginal distribution problem

1. Jun 4, 2007

### FunkyDwarf

Hey guys,

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

1. The problem statement, all variables and given/known data
Find fx(x,y) of
$$f(x,y) = \frac{(1+4xy)}{2} for 0 \leq x, y \leq 1$$ and zero otherwise

2. Relevant equations
Now this should equal$$\int \frac{(1+4xy)}{2} dy$$over all y but that leads to infinities ( as y goes from minus infinity to 1)which obviously we can't have. Im sure im 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

Last edited: Jun 4, 2007
2. Jun 4, 2007

### NateTG

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

3. Jun 4, 2007

### 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.

4. Jun 4, 2007

### 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.

5. Jun 4, 2007

### NateTG

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

Last edited: Jun 4, 2007
6. Jun 4, 2007

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.

7. Jun 4, 2007

### ZioX

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

8. Jun 5, 2007

### FunkyDwarf

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