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Thread: Expected value of Bivariate distribution

  1. #1
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    Expected value of Bivariate distribution

    I'm a little stuck on this question:
    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    e^{-(y_1+y_2)} & \mbox{for} & y_1 > 0, \ \ y_2 >0 \\
    0 & \mbox{for} & \mbox{otherwise}
    \end{array}\right.$

    a) What are $\displaystyle E(Y_1+Y_2)$ and $\displaystyle V(Y_1+Y_2)$ ?

    b) What is $\displaystyle P(Y_1-Y_2>3)$ ?

    so far I have:

    a) $\displaystyle E(Y_1+Y_2) = E(Y_1) +E(Y_2)$

    $\displaystyle E(Y_1) = \int_0^{\infty} y_1e^{-(y_1+y_2)} \ dy_1$

    $\displaystyle = y_1e^{-(y_1+y_2)} - (-e^{-(y_1+y_2)}) \bigg{|}_0^{\infty}$

    $\displaystyle = (0 + 0) - (0 -e^{-y_2})= e^{-y_2}$

    $\displaystyle E(Y_2) = e^{-{y_1}}$

    $\displaystyle E({Y_1}^2) = \int_0^{\infty} {y_1}^2 e^{-(y_1+y_2)} \ dy_1$

    integration by parts gives me $\displaystyle =-{y_1}^2e^{-(y_1+y_2)} +2y_1 e^{-(y_1+y_2)}- 2e^{-(y_1+y_2)} \bigg{|}_{0}^{\infty} $

    $\displaystyle = e^{-(y_1+y_2)}(-{y_1}^2+2y_1-2)\bigg{|}_{0}^{\infty}$

    $\displaystyle = 0 - e^{-y_2}(2) = 2e^{-y_2}$

    $\displaystyle E({Y_2}^2) = 2e^{-y_1}$

    now to get the variance, it's $\displaystyle [E({Y_2}^2) +E({Y_1}^2)] - [(E({Y_1+Y_2}))^2] $

    $\displaystyle = 2e^{-y_1} + 2e^{-y_2} - (e^{-y_1}+e^{-y_2})^2$

    At this point I know I made a mistake since the answer in the back of the book is 2 and I got nowhere near that.


    b) $\displaystyle y_1 >3 \ \ \mbox{and} \ \ 0 < y_2 < y_1-3$

    $\displaystyle = \int_3^{\infty} \int_0^{y_1-3} e^{-(y_1+y_2)} \ dy_2\ dy_1$

    $\displaystyle = \int_3^{\infty} \left( -e^{-(y_1+y_2)} \right) \bigg{|}_0^{y_1-3} \ dy_1$

    $\displaystyle = \int_3^{\infty} \left( -e^{-(y_1+(y_1-3))} - (-e^{-(y_1+0)}) \right) \ dy_1$

    $\displaystyle = \int_3^{\infty} \left( -e^{-(2y_1-3)} + e^{-y_1} \right) \ dy_1$

    $\displaystyle = \int_3^{\infty} -e^{-(2y_1-3)} \ dy_1 + \int_3^{\infty} e^{-y_1} \ dy_1$

    $\displaystyle = \frac{1}{2}e^{-(2y_1-3)}\bigg{|}_3^{\infty} - e^{-y_1} \bigg{|}_3^{\infty}$

    = $\displaystyle (0 -\frac{1}{2}e^{-3}) + (0+e^{-3}) = \frac{1}{2}e^{-3}$
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lllll View Post
    I'm a little stuck on this question:
    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    e^{-(y_1+y_2)} & \mbox{for} & y_1 > 0, \ \ y_2 >0 \\
    0 & \mbox{for} & \mbox{otherwise}
    \end{array}\right.$
    $\displaystyle Y_1$ and $\displaystyle Y_2$ are independent, with densities $\displaystyle e^{-y_1}$ and $\displaystyle e^{-y_2}$ , so:

    $\displaystyle E(Y_1)=\int_0^{\infty}y_1 e^{-y_1} \ dy_1$

    and so on

    RonL
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