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Math Help - Expected value of the product of two rvs

  1. #1
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    Expected value of the product of two rvs

    Given two r.v.s Y1 and Y2 that have joint density fcn:

    f(y_1,y_2) = [1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_1-y_2} for 0 <= y_1, 0 <=y_2,
    and 0 elsewhere

    and -1 \le \alpha \le 1

    I have to find E(Y_1Y_2).

    I did this by doing:
    \int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2[1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_2}\,dy_2] \,dy_1

    But I end up at a different answer than the final answer in the back of my text. The answer given in the back is:
    E(Y_1Y_2) = 1 - \alpha/4
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  2. #2
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    Quote Originally Posted by Canadian0469 View Post
    Given two r.v.s Y1 and Y2 that have joint density fcn:

    f(y_1,y_2) = [1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_1-y_2} for 0 <= y_1, 0 <=y_2,
    and 0 elsewhere

    and -1 \le \alpha \le 1

    I have to find E(Y_1Y_2).

    I did this by doing:
    \int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2[1-\alpha[(1-2e^{-y_1})(1-2e^{-y_2})]]e^{-y_2}\,dy_2] \,dy_1

    But I end up at a different answer than the final answer in the back of my text. The answer given in the back is:
    E(Y_1Y_2) = 1 - \alpha/4
    I agree with the answer 1 - \frac{\alpha}{4}.

    Stage 1:

    I get \int^{+\infty}_0 y_2 \left[ 1 - \alpha (1-2e^{-y_1}) (1-2e^{-y_2}) \right] e^{-y_2} \,dy_2 = \frac{1}{2} \left( 2 e^{y_1} - \alpha \left[ e^{y_1} - 2\right] \right) \cdot e^{-y_1}.

    Do you?
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  3. #3
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    No, I can't see where you got that.
    I started off by expanding the inner expression (the integration with respect to y_2, and then worked with
    <br />
\int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2e^{-y_2} - \alpha y_2e^{-y_2} + 2\alpha y_2 e^{-2y_2} + 2\alpha y_2 e^{-y_1-y_2} - 4\alpha y_2 e^{-y_1-2y_2} \,dy_2] \,dy_1<br />

    This was the only approach I could think of... Where did I go wrong?
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  4. #4
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    Quote Originally Posted by Canadian0469 View Post
    No, I can't see where you got that.
    I started off by expanding the inner expression (the integration with respect to y_2, and then worked with
    <br />
\int^\infty_0 y_1e^{-y_1}[\int^\infty_0y_2e^{-y_2} - \alpha y_2e^{-y_2} + 2\alpha y_2 e^{-2y_2} + 2\alpha y_2 e^{-y_1-y_2} - 4\alpha y_2 e^{-y_1-2y_2} \,dy_2] \,dy_1<br />

    This was the only approach I could think of... Where did I go wrong?
    The answer I've given for the 'inner integral' is correct. You then use this result to do the 'outer integration'.

    I have no idea where you're going wrong - to know that you will need to post all your working.
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  5. #5
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    Okay I have the same result for the inner integral now. I'll continue tonight and will post the result. Thanks for your support mr. fantastic!
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  6. #6
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    Okay, I finally got the correct answer. I really had to break it up the way you did to get it. It took less time than doing it the way I was doing it before (expanding the whole expression). I suppose I was making some sort of algebraic mistake before?

    Thanks mr. fantastic, you are awesome!
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