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Homework Help: Independence of Random Variables

  1. Dec 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Given ##f_{X,Y}(x,y)=2e^{-x}e^{-y}\ ;\ 0<x<y\ ;\ y>0##,
    The following theorem given in my book (Larsen and Marx) doesn't appear to hold.

    2. Relevant equations
    Definition
    ##X## and ##Y## are independent if for every interval ##A## and ##B##, ##P(X\in A \land Y\in B) = P(X\in A)P(Y\in B) ##.
    Theorem
    ##X## and ##Y## are independent iff ##f_{X,Y}(x,y)=g(x)h(y)##.
    If so, there is a constant ##k## such that ##f_X(x)=kg(x)## and ##f_Y(y)=(1/k)h(y)##.

    3. The attempt at a solution
    Consider ##g(x)=2e^{-x}## and ##h(y)=e^{-y}##. Then, ##f_{X,Y}(x,y)=g(x)h(y)##, therefore theorem indicates that ##X## and ##Y## are independent.
    The constant ##k## is
    ##k=\int_0^\infty h(y)dy=1##
    ##k=\int_0^y g(x)dx = 2(1-e^{-y})##
    There is a contradiction in the value of k and it is not constant.

    Am I missing something, or is the theorem incomplete in that it is lacking details on the intervals that the random variables are defined on?
     
  2. jcsd
  3. Dec 2, 2017 #2

    MathematicalPhysicist

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    The theorem is correct, you should have: ##f_X(x) = \int_x^\infty f_{X,Y}(x,y)dy## and ##f_Y(y) = \int_0^y f_{X,Y}(x,y)dx##.
     
  4. Dec 2, 2017 #3

    Ray Vickson

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    You are definitely missing something: your "product" formula for ##f_{X,y}(x,y) ## holds only over the region ##0 < x < y##. (Presumably, ##f_{X,Y}## is zero outside that region, but the question does not say that.) Anyway, a bivariate density for a pair of independent random variables would need to be non-zero on the whole of the bivariate space ## 0 < x,y < \infty##.

    To see explicitely that ##X,Y## are NOT independent, compute the marginal densities
    $$ f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy = \int_x^{\infty} e^{-x} e^{-y} \, dy$$
    and
    $$f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dx =\int_0^y e^{-x} e^{-y} \, dx .$$
    Finish the calculations and then check if ##f_X(x) f_Y(y)## agrees with your ##f_{X,Y} (x,y)##
     
    Last edited: Dec 2, 2017
  5. Dec 2, 2017 #4
    ##f_X(x) = 2\int_x^\infty e^{-x}e^{-y}dy = 2e^{-2x}##
    ##f_Y(y) = 2\int_0^y e^{-x}e^{-y}dx = 2e^{-y}(1-e^{-y})##
    ##f_X(x)f_Y(y) = 4e^{-2x}e^{-y}(1-e^{-y}) \neq f_{X,Y}(x,y)##
    So we can show explicitly that they are not independent...
    But we can separate ##f_{X,Y}(x,y)## into a product of ##g(x)## and ##h(y)## which implies independence by the theorem?
     
  6. Dec 2, 2017 #5

    Ray Vickson

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    No, no, no! Absolutely not! Your ##f_{X,Y}(x,y)## is NOT a product of some ##g(x)## and some ##h(y)## over the whole quarter plane ##0 < x, y < \infty##. It is a product, but only over the restricted region ##0 < x < y < \infty##, and that fact makes ##X## and ##Y## very definitely not independent. In fact, you should do what I suggested in my first response, which is to compute the marginals ##f_X(x)## and ##f_Y(y)##. You can use these to compute ##P(X \in A)## and ##P(Y \in B)## for ##A = \{ 0 < x < 1 \}## and ##B= \{ 0 < y < 2 \}##. Now compute ##P(X \in A) \cdot P(Y \in B)##. Finally, compute ##P(X \in A \; \& \; Y \in B)## by integrating ##f_{X,Y}(x,y)## over the appropriate 2-dimensional region. Do you get different answers? (I do.)
     
  7. Dec 2, 2017 #6
    Yes, the answers are different. Can ##X## and ##Y## only be independent if they are defined over a rectangle?
     
  8. Dec 2, 2017 #7

    Ray Vickson

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    Basically, yes, and for obvious reasons: you cannot make ##g(x) h(y)## come out at zero over a part of the ##(x,y)## plane where both ##g(x) \neq 0## and ##h(y) \neq 0##.
     
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