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Finding marginal density functions

  1. Dec 9, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    The joint probability density function of ##X## and ##Y## is given by $$f(x,y) = \frac{1}{8}(y^2 - x^2)e^{-y},\,\,\,\, x \in\,[-y,y]\,\,,y \in\,(0, \infty)$$

    Compute the marginal densities of ##X## and ##Y##.

    3. The attempt at a solution
    I know the defintions are $$ F_X(x) = \int_{- \infty}^{\infty}\,f(x,y)\,dy\, \text{and}\, F_Y(y) = \int_{- \infty}^{\infty}\,f(x,y)\,dx.$$

    Am I correct in saying that if the domains of x and y are just numbers then the limits on the integrals are the endpoints of these domains? I.e if in the example above ##x \in\, [2,4], \,\,\text{and}\,\, y\,\in\,[1,3] F_Y(y) = \int_2^4 f(x,y)\,dx,\,\,\text{while}\,\, F_X(x) = \int_1^3 f(x,y)\,dy##, right?

    However, in this case, the domain of x depends on y. So x is less than y=x and greater than y = -x. Drawing this, I see it is the portion enclosed by x = |y|, but above the x axis (since y is greater than 0). Hence, to get the limits when computing ##F_X(x)##, I should say y is from x to infinity. When I check the answers, they have y from |x| to infinity. Can't I replace |x| = x since y is above 0? Or did I miss something else? Many thanks
     
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  3. Dec 9, 2012 #2

    Ray Vickson

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    Two points.
    (1) You should use the notations ##f_X(x)## instead of ##F_X(x)##, etc., because F looks like a cumulative distribution function. In probability is is common to use small letters for densities or probability mass functions and capital letters for (cumulative) distribution functions.
    (2) Always draw a picture when evaluating 2D integrals. If you draw the allowable combinations of x and y, everything will become clear.
     
  4. Dec 9, 2012 #3

    CAF123

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    I forgot the portion on the left hand side - it's clear now. I have a conceptual question: what does a marginal density actually mean?
     
  5. Dec 9, 2012 #4

    Ray Vickson

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    The marginal density of X is just the probability density of X. The random variable X may be one component of a much more extensive list (or vector) of properties, but we are just ignoring those other components (if any) when we look at the marginal. In your case, X is the first component of a vector (X,Y), but when we look at the marginal density of X we are ignoring Y. Isn't all this explained in your textbook? If not, I am very surprised.
     
  6. Dec 10, 2012 #5

    CAF123

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    I see. My book did contain that, however it is more geared toward giving examples rather than explaining the theory so that might be why I missed it first time. From this same question, I asks to compute ##E[X]##. I know this is equal to ##\int_{-∞}^{∞} x f_X(x) dx##, so can I write this as ##\int_{-y}^y x f_X(x) dx##. In the book, they take the limits of integration as 0 to ∞. Why is this? I know x is from -y to y, so can't x take on negative values too? (If y = 5, say, then x is from -5 to 5?). It would seem to me that the limits be from -∞ to ∞?
     
  7. Dec 10, 2012 #6

    Ray Vickson

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    The two integrals
    [tex]\int_{-∞}^{∞} x f_X(x) dx [/tex]
    and
    [tex]\int_{-y}^y x f_X(x) dx [/tex]
    are not equal. Why would you think they are the same?
     
  8. Dec 10, 2012 #7

    CAF123

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    The former integral is if x is not restricted to any domain and the latter if it is. No?
     
  9. Dec 10, 2012 #8

    Ray Vickson

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    No, not in this problem. When you talk about f_X there is no y anywhere; y is gone.
     
  10. Dec 10, 2012 #9

    CAF123

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    Oh, pardon me, what a mistake to make. The expectation should be a number, not a function. But why are the limits from 0 to ∞?
     
  11. Dec 10, 2012 #10

    Ray Vickson

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    I said it before, and I will say it again one more time: draw a diagram! This will show the region in the (x,y) plane where f_{XY}(x,y) > 0.
     
  12. Dec 10, 2012 #11

    CAF123

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    I did draw a diagram but the region -y < x < y is the region above y = |x| in the xy plane. That's why I think the limits are from -∞ to ∞ for x. I don't know why x is resticted from 0 to ∞. To get that case, we would need x > |y| I think.
     
  13. Dec 10, 2012 #12

    Ray Vickson

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    The function f_X(x) = f(x) is symmetric, so the integral of x*f(x) from -inf to +inf is zero; that is, EX = 0. The integral of x*f from 0 to +inf is positive, so it looks like they are computing (1/2) E|X|.
     
  14. Dec 10, 2012 #13

    CAF123

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    The answer in the back of the book is that EX = 0. Does this mean that the limits are actually -∞ to ∞? When I check the solutions, they compute the integral from 0 to ∞, yet still get 0 (When I work it through, I get 3/4, which as you said is indeed positive).
     
  15. Dec 10, 2012 #14

    Ray Vickson

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    You can have EX = 0 without having limits of -∞ and +∞---that is a whole separate issue. The question is: what are the limits in THIS problem? Well, can you have values of x > 100? can you have x < -1000? x > 100,000,000? Is there ANY M > 0 that has P{X > M} = 0 and P(X < -M} = 0? If you think there is, please tell me a suitable value of M. If there is no such value of M, what is that telling you? (I am sure you already know the answer; it just seems to be the case that you lack confidence in your own answers.)
     
  16. Dec 11, 2012 #15

    CAF123

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    There is no such M. As long as x lies between -y and y. And since y is from 0 to infinity, the value of x can take any value in (-∞, ∞).
     
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