Simple marginal distribution problem

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Homework Help Overview

The discussion revolves around finding the marginal distribution function from a given joint probability density function (pdf) defined as f(x,y) = (1 + 4xy)/2 for the domain 0 ≤ x, y ≤ 1. Participants are exploring the implications of the defined domain and the behavior of the function outside this range.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to properly define the domain of the function and the implications of integrating over an incorrect range. There are suggestions to split the integral and to visualize the support of the function to clarify the setup.

Discussion Status

The conversation is active, with participants providing guidance on understanding the domain of the function and the importance of ensuring the pdf integrates to 1. There is recognition of the need to clarify assumptions about the variables involved.

Contextual Notes

There is an emphasis on the constraints of the problem, particularly regarding the defined domain of the function and the implications of integrating over an incorrect range. Participants note that the function is non-zero only within the specified limits.

FunkyDwarf
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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
[tex]f(x,y) = \frac{(1+4xy)}{2} for 0 \leq x, y \leq 1[/tex] and zero otherwise

Homework Equations


Now this should equal[tex]\int \frac{(1+4xy)}{2} dy[/tex]over 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
 
Last edited:
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You might want to split the integral into a couple of pieces. Remeber:
[tex]f(x,y) \neq \frac{1+4xy}{2}[/tex]
 
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 dependent on each other...which can make things complicated.
 
Doesn't this expression,

[tex]0 \leq x, y \leq 1[/tex]

mean x and y are both in [0,1]? If y runs to minus infinity the question doesn't make much sense.
 
ZioX said:
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, [itex]f(x,y)[/itex] is only non-zero on a certain domain, it's defined on the entire plane.
 
Last edited:
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.
 
NateTG said:
Actually, [itex]f(x,y)[/itex] 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.
 
Ah of course! i did draw it, i just drew it wrong :P thanks guys
 

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