Random sample $Y_1,\dots, Y_n$ of size n from a univariate normal population with ($\mu, \sigma^2$).
Let $\bar{y}=\frac{1}{n}\sum Y_i$.
$\sum(y_i-\bar{y})^2$ can be written in the for $\sigma^2 X'AX$ where $X\sim N(0,1)$. What is $A$?
My solution:
Since $(y_i-\bar{y}) \sim N(0,\frac{n+1}{n} \sigma^2)$, $\frac{n}{(n+1)\sigma^2}\sum(y_i-\bar{y})^2 \sim \chi^2(n)$.
$X'X \sim \chi^2(1)$. I stuck here...
Assuming that $(Y_i)$ is independent, $Y_i=\mu+\sigma X_i$ for every $i$, where $X=(X_i)$ is i.i.d. standard normal. Thus, $\bar Y=\frac1n\sum\limits_{i=1}^nY_i$ is $\bar Y=\mu+\frac1n\sigma U^TX$, where $U$ is the vector whose every entry is $1$.
For every $i$, $Y_i=\mu+\sigma E_i^TX$ where $E_i$ is the vector whose only nonzero entry is $1$ at the $i$th place, hence $Y_i-\bar Y=\sigma V_i^TX$, where $V_i=E_i-\frac1nU$, and $(Y_i-\bar Y)^2=(Y_i-\bar Y)^T(Y_i-\bar Y)$ hence $(Y_i-\bar Y)^2=\sigma^2 X^TV_iV_i^TX$. One sees that $S^2=\sum\limits_{i=1}^n(Y_i-\bar Y)^2$ is $S^2=\sigma^2X^TAX$, with $$ A=\sum_{i=1}^nV_iV_i^T=\sum_{i=1}^n(E_i-\tfrac1nU)(E_i-\tfrac1nU)^T=F-\tfrac1nEU^T-\tfrac1nUE^T+\tfrac1nUU^T, $$ where $F=\sum\limits_{i=1}^nE_iE_i^T$ and $E=\sum\limits_{i=1}^nE_i$ hence $E=U$. Each $E_iE_i^T$ is the matrix whose only nonzero coefficient is $1$ at the $(i,i)$ place hence $F=I_n$ is the identity matrix. And $UU^T=W$ is the square matrix whose every coefficient is $1$. Thus, $$ A=I_n-\tfrac1nW. $$