1. ## E[W] f distribution

Let

$\displaystyle W = \frac{{\sum\limits_{i = 1}^{16} {(X_i - 10)^2 } }}{{\sum\limits_{j = 1}^{10} {(Y_j - 15)^2 } }}$
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]?

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

My Question:
Why 120/64?

I always get 64/120.

Any help is greatly appreciated.

2. Originally Posted by kid funky fried
Let

$\displaystyle W = \frac{{\sum\limits_{i = 1}^{16} {(X_i - 10)^2 } }}{{\sum\limits_{j = 1}^{10} {(Y_j - 15)^2 } }}$
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]?

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

My Question:
Why 120/64?

I always get 64/120.

Any help is greatly appreciated.
$\displaystyle E[W] = E[\frac{{\sum\limits_{i = 1}^{16} {(X_i - 10)^2 } }}{{\sum\limits_{j = 1}^{10} (Y_j - 15)^2 }}] = \frac{{64}}{{120}}E[\frac{\frac{\sum\limits_{i = 1}^{16} {(X_i - 10)^2 } }{4\cdot 16}}{\frac{\sum\limits_{j = 1}^{10} (Y_j - 15)^2 }{12\cdot 10}}] = \frac{{64}}{{120}}\cdot\frac{{1}}{1}$
since $\displaystyle E[(X_i-10)^2]={\rm Var}(X_i)=4$ and the same for $\displaystyle Y_j$.