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Thread: continous random variable

  1. #1
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    continous random variable

    Y has a distribution function:

    $\displaystyle F(y) = \left\{ \begin{array}{rcl} 0 & \mbox{for} & y \leq 0 \\
    \\
    \frac{y}{8} & \mbox{for} & 0 < y < 2 \\
    \\
    \frac{y^2}{16} & \mbox{for} & 2 \leq y < 4 \\
    \\
    1 & \mbox{for} & y \geq y 4
    \end{array}\right.$

    find $\displaystyle \mu$ and $\displaystyle \sigma^2$

    since $\displaystyle \mu = \int yf(y) dy \therefore \int^{2}_{0} y \left( \frac{y}{8} \right) dy + \int^{4}_{2} y \left( \frac{y^2}{16} \right) dy + \int^{\infty}_{4} y(1) dy= \frac{y^3}{24} \bigg{|}^{2}_{0} + \ \ \frac{y^4}{64} \bigg{|}^{4}_{2} + \ \ \frac{y^2}{2} \bigg{|}_{4}$ $\displaystyle = \frac{1}{3} +\frac{256-16}{16} + 8 = \frac{145}{12}$

    I know that my final solution is incorect since it's suppossed to be $\displaystyle \frac{31}{12}$, where did I go wrong?
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  2. #2
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    differentiate $\displaystyle F(y) $ to get $\displaystyle f(y) $ (not integrate). Then take $\displaystyle \int yf(y) \ dy $ to get $\displaystyle \mu $.
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