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Math Help - Covariance - Joint PDF

  1. #1
    Junior Member utopiaNow's Avatar
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    Covariance - Joint PDF

    Given:
    <br />
f(x, y) = xy^2exp(-(x+1)y) \ x, y > 0<br />

    I needed to calculate the Cov(X,Y).

    So I used the formula Cov(X, Y) = E(XY) - E(X)E(Y)

    The problem occurs when I go to calculate E(X).

    I know  E(X) = \int_o^\infty \! xf_x(x) \, dx.

    Where the marginal pdf of x is  f_x(x) = \frac{2x}{(1 + x)^3} .
    So the integral now is:
    <br />
\int_o^\infty \! \frac{2x^2}{(1 + x)^3}  \, dx.<br />

    Now when I evaluate that integral I get a divergent integral evaluating to  \infty . I've checked my work several times, but I can't seem to find what I'm doing wrong. Any suggestions?

    Thanks in advance.
    Last edited by utopiaNow; April 6th 2009 at 11:40 AM.
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  2. #2
    Moo
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    Hello,

    Now when I evaluate that integral I get an improper integral. I've checked my work several times, but I can't seem to find what I'm doing wrong. Any suggestions?
    Well, what is an improper integral to you ?
    Because it surely is an improper integral here, caused by the infinity boundary !

    Now, I get the same problems as yours...

    Setting \int_0^\infty \frac{2x^2}{(1+x)^3} ~dx=\lim_{a \to \infty} \int_0^a \frac{2x^2}{(1+x)^3} ~dx, I get :

    \int_0^a \frac{2x^2}{(1+x)^3} ~dx=6 \left(2\ln(1+a)-2 \cdot \frac{a}{1+a}-\frac{a^2}{(1+a)^2}\right)


    And we also can see that \frac{2x^2}{(1+x)^3} \sim \frac 2x as x is very big.
    And \int_1^\infty \frac 1x ~dx is a divergent Riemann integral !




    My guess is that you should not calculate separately the expectations (you should know that you don't necessarily have \int f+g=\int f+\int g, especially if there are integrals that don't exist), by using the \text{Cov}(X,Y)=\mathbb{E}\{(X-\mathbb{E}(X))(Y-\mathbb{E}(Y))\}

    We proved in our class that if X and Y are bounded, then :
    \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)=\int_{\mathbb{R}} \int_{\mathbb{R}} \left[\mathbb{P}(X\leq x,Y\leq y)-\mathbb{P}(X\leq x)\mathbb{P}(Y\leq y)\right] ~dxdy

    But I must say I don't know if it is the correct way...
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  3. #3
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by Moo View Post
    Well, what is an improper integral to you ?
    Because it surely is an improper integral here, caused by the infinity boundary !
    Oops sorry I meant divergent. I'll try what you suggested and see if it works.
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    I'm doing the same question.

    I tried calculating E(X), and after integrating, I found

    <br />
\frac{4x+3}{(x+1)^2} + 2log(x+1)<br />

    but with the limits of integration being 0 to infinity, I stopped there, since the log(x) diverges to infinity and I get moot.

    I thought that maybe after finding E_{Y}(Y) that I could use

    <br />
E_{X}(X) = E_{Y}(E_{X \mid Y}(X \mid Y=y))<br />

    with  E_{X \mid Y}(X \mid Y=y) = \frac{2}{y}, (if I did my calculations correctly) but again... the integral gets weird. I'm all out of ideas...
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    Junior Member utopiaNow's Avatar
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    Yeah I ran into both those problems as well. And now I am also officially out of ideas. I think I'll have to ask the teacher tomorrow.
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    Hmm... looks like the question changed.

    For Q1(d), change the question to "Does the Cov(X,Y) exist? Show work to justify your answer."
    I'd say it doesn't exist
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  7. #7
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by veol View Post
    Hmm... looks like the question changed.

    [color=#FF0000]

    I'd say it doesn't exist
    Woot good news. Thanks for that! I was still struggling with that one. LOL.
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    i am doing the same question...but i dont know how to do part e) find explicit expression of MGF of ln Y...and specify the range of t...
    anyone can help me? Thanks a lot!
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    Quote Originally Posted by cm917 View Post
    i am doing the same question...but i dont know how to do part e) find explicit expression of MGF of ln Y...and specify the range of t...
    anyone can help me? Thanks a lot!
    The hard way: Get the marginal of Y. Find the pdf of ln Y. Apply the definition of MGF. Do the integration.

    The easy way: Get the marginal of Y. Note that E\left( e^{t \ln Y} \right) = E\left( Y^t\right). Do the integration.
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    Thanks! but i dont know how to find the pdf of lny...Would you please show it to me ? Thnaks a lot!
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  11. #11
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    Quote Originally Posted by cm917 View Post
    Thanks! but i dont know how to find the pdf of lny[snip]
    Then do it the other way.
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    i tired the easy way...
    E[y^t) = Integrate from 0 to infinity...( y^t) (e^-y) dy....
    i applied the part rule... then I need to integrate (e^-y ) (y^(t-1)) dy again...it seems even i keep integrate it..i cant get rid of t...did i do something wrong?
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  13. #13
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    Quote Originally Posted by cm917 View Post
    i tired the easy way...
    E[y^t) = Integrate from 0 to infinity...( y^t) (e^-y) dy....
    i applied the part rule... then I need to integrate (e^-y ) (y^(t-1)) dy again...it seems even i keep integrate it..i cant get rid of t...did i do something wrong?
    Use the Gamma function: Gamma function - Wikipedia, the free encyclopedia
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