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Math Help - clearification on probability

  1. #1
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    clearification on probability

    find E[y_1].

    f(y_1,y_2) = \left\{ \begin{array}{rcl}<br />
\frac{1}{8}y_1e^{-(y_1+y_2)/2} & \mbox{if} & y_1>0, \ y_2>0 3 \\ <br />
0 & \mbox{if} & \mbox{otherwise} \end{array}\right.<br />

    E[y_1] = \int_0^\infty \int_0^\infty \frac{1}{8}y_1^2e^{-(y_1+y_2)/2} \  dy_2 \ dy_1

    E[y_1] = \int_0^\infty \frac{1}{8}y_1^2e^{-y_1/2} \int_0^\infty e^{-y_2/2} \  dy_2 \ dy_1

    E[y_1] = \int_0^\infty \frac{1}{8}y_1^2e^{-y_1/2} \bigg{[}-2e^{-y_2/2}\bigg{]}^\infty_0 \ dy_1

    E[y_1] = \frac{1}{4}\int_0^\infty y_1^2e^{-y_1/2} \ dy_1

    E[y_1] = \frac{1}{4}\bigg{[} -2y_1^2e^{-y_1/2} -[-2y_1e^{-y_1/2} -(-2e^{-y_1/2})] \bigg{]}^\infty_0

    E[y_1] = \frac{1}{4}\bigg{[} -2y_1^2e^{-y_1/2} +2y_1e^{-y_1/2} -2e^{-y_1/2} \bigg{]}^\infty_0

    E[y_1] = {\color{blue}-\frac{1}{2}y_1^2e^{-y_1/2} \bigg{|}^\infty_0}+{\color{blue}\frac{1}{2}y_1e^{-y_1/2}\bigg{|}^\infty_0} -\frac{1}{2}e^{-y_1/2}\bigg{|}^\infty_0

    now what I'm not so sure about are the expressions that are in blue. Will they give:

    1) {\color{blue} -(\infty-0)}+{\color{blue}\infty+0} +\frac{1}{2} = \frac{1}{2}

    or will it be:

    2) {\color{blue}-1}+{\color{blue}1} +\frac{1}{2} = \frac{1}{2}

    I would think that it's the 2) since the integral from 0 to infinity is just equal to on, but 1) also seems right for some reason.
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  2. #2
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    Quote Originally Posted by lllll View Post
    find E[y_1].

    f(y_1,y_2) = \left\{ \begin{array}{rcl}<br />
\frac{1}{8}y_1e^{-(y_1+y_2)/2} & \mbox{if} & y_1>0, \ y_2>0 3 \\ <br />
0 & \mbox{if} & \mbox{otherwise} \end{array}\right.<br />

    E[y_1] = \int_0^\infty \int_0^\infty \frac{1}{8}y_1^2e^{-(y_1+y_2)/2} \ dy_2 \ dy_1

    E[y_1] = \int_0^\infty \frac{1}{8}y_1^2e^{-y_1/2} \int_0^\infty e^{-y_2/2} \ dy_2 \ dy_1

    E[y_1] = \int_0^\infty \frac{1}{8}y_1^2e^{-y_1/2} \bigg{[}-2e^{-y_2/2}\bigg{]}^\infty_0 \ dy_1

    E[y_1] = \frac{1}{4}\int_0^\infty y_1^2e^{-y_1/2} \ dy_1

    E[y_1] = \frac{1}{4}\bigg{[} -2y_1^2e^{-y_1/2} -[-2y_1e^{-y_1/2} -(-2e^{-y_1/2})] \bigg{]}^\infty_0

    E[y_1] = \frac{1}{4}\bigg{[} -2y_1^2e^{-y_1/2} +2y_1e^{-y_1/2} -2e^{-y_1/2} \bigg{]}^\infty_0

    E[y_1] = {\color{blue}-\frac{1}{2}y_1^2e^{-y_1/2} \bigg{|}^\infty_0}+{\color{blue}\frac{1}{2}y_1e^{-y_1/2}\bigg{|}^\infty_0} -\frac{1}{2}e^{-y_1/2}\bigg{|}^\infty_0 *

    now what I'm not so sure about are the expressions that are in blue. Will they give:

    1) {\color{blue} -(\infty-0)}+{\color{blue}\infty+0} +\frac{1}{2} = \frac{1}{2}

    or will it be:

    2) {\color{blue}-1}+{\color{blue}1} +\frac{1}{2} = \frac{1}{2}

    I would think that it's the 2) since the integral from 0 to infinity is just equal to on, but 1) also seems right for some reason.
    I'll assume everything up to and including the line marked * is correct (I haven't checked).

    Then it's actually

    3) (0 - 0) + (0 - 0) - (0 - [1/2]) = 1/2. Same answer as 2) but the working in 2) is incorrect.
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