Originally Posted by

**noob mathematician** Oh, the answers I gave were pure guessing, however i hope to know the explantion: I was stuck half way through my working as followed.

b)Find the distribution of $\displaystyle U=\sum\limits_{i=1}^5 (Y_i-\overline{Y})^2$.

I actually expanded the question:

since $\displaystyle \sum\limits_{i=1}^n (X_i-\overline{X})^2$

=$\displaystyle \sum\limits_{i=1}^n (X_i^2-2X_i\overline{X}+\overline{X}^2)$

=$\displaystyle \sum\limits_{i=1}^nX_i^2-2n\overline{X}^2+n\overline{X}^2$

=$\displaystyle \sum\limits_{i=1}^nX_i^2-n\overline{X}^2$

Applying to my question will yield:

$\displaystyle \sum\limits_{i=1}^5Y_i^2-5\overline{Y}^2$,

then since it was defined initially that $\displaystyle \overline{Y}=\frac{1}{5}\sum\limits_{i=1}^5 Y_i$, then it should have a distribution of N(0,1) too.

Subsequently, I will have: $\displaystyle U={\chi_5}^2-5(Z)^2={\chi_5}^2-Z_1^2-Z_2^2-Z_3^2-Z_4^2-Z_5^2$, where Z is the standard normal. Isn't the equation just cancelled out each other? How does it still remain as $\displaystyle {\chi_5}^2$?