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Math Help - E[W] f distribution

  1. #1
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    E[W] f distribution

    Let


    <br />
W = \frac{{\sum\limits_{i = 1}^{16} {(X_i  - 10)^2 } }}{{\sum\limits_{j = 1}^{10} {(Y_j  - 15)^2 } }}<br />
    where X's be i.i.d. Normal with mean 10 and variance 4 for i=1,2,...,16 and Y's be i.i.d. Normal with mean 15 and variance 12 for j=1,2,...10 and X and Y are independent for all i and j. What is the value of E[W]?

    Professor's answer:


    <br />
E[W] = E[\frac{{\sum\limits_{i = 1}^{16} {(X_i  - 10)^2 } }}{{\sum\limits_{j = 1}^{10} {(Y_j  - 15)^2 } }}] = \frac{{120}}{{64}}E[\frac{{\frac{{\frac{{\sum\limits_{i = 1}^{16} {(X_i  - 10)^2 } }}{4}}}{{16}}}}{{\frac{{\frac{{\sum\limits_{j = 1}^{10} {(Y_j  - 15)^2 } }}{{12}}}}{{10}}}}] = \frac{{120}}{{64}}*\frac{{10}}{8} = 2.34375<br />

    My Question:
    Why 120/64?

    I always get 64/120.

    Any help is greatly appreciated.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by kid funky fried View Post
    Let


    <br />
W = \frac{{\sum\limits_{i = 1}^{16} {(X_i  - 10)^2 } }}{{\sum\limits_{j = 1}^{10} {(Y_j  - 15)^2 } }}<br />
    where X's be i.i.d. Normal with mean 10 and variance 4 for i=1,2,...,16 and Y's be i.i.d. Normal with mean 15 and variance 12 for j=1,2,...10 and X and Y are independent for all i and j. What is the value of E[W]?

    Professor's answer:


    <br />
E[W] = E[\frac{{\sum\limits_{i = 1}^{16} {(X_i  - 10)^2 } }}{{\sum\limits_{j = 1}^{10} {(Y_j  - 15)^2 } }}] = \frac{{120}}{{64}}E[\frac{{\frac{{\frac{{\sum\limits_{i = 1}^{16} {(X_i  - 10)^2 } }}{4}}}{{16}}}}{{\frac{{\frac{{\sum\limits_{j = 1}^{10} {(Y_j  - 15)^2 } }}{{12}}}}{{10}}}}] = \frac{{120}}{{64}}*\frac{{10}}{8} = 2.34375<br />

    My Question:
    Why 120/64?

    I always get 64/120.

    Any help is greatly appreciated.
    The answer should be:

    <br />
E[W] = E[\frac{{\sum\limits_{i = 1}^{16} {(X_i - 10)^2 } }}{{\sum\limits_{j = 1}^{10} (Y_j - 15)^2  }}] = \frac{{64}}{{120}}E[\frac{\frac{\sum\limits_{i = 1}^{16} {(X_i - 10)^2 } }{4\cdot 16}}{\frac{\sum\limits_{j = 1}^{10} (Y_j - 15)^2  }{12\cdot 10}}] = \frac{{64}}{{120}}\cdot\frac{{1}}{1} <br />

    since E[(X_i-10)^2]={\rm Var}(X_i)=4 and the same for Y_j.
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