# Assume that $Z_i\sim\mathcal{N}(0,1)$ are independent and prove that $\sum\limits_{i=1}^{n}(Z_i-\overline{Z})^{2}\sim\chi^{2}_{(n)}$

Assume that $$Z_{i}\sim\mathcal{N}(0,1)$$ are independent and prove the following results

(a) $$\displaystyle\overline{Z} = \frac{1}{n}\sum_{i=1}^{n}Z_{i}\sim\mathcal{N}(0,1/n)$$

(b) $$\overline{Z}$$ and $$\displaystyle\sum_{i=1}^{n}(Z_{i} - \overline{Z})^{2}$$ are independent random variables.

(c) $$\displaystyle\sum_{i=1}^{n}(Z_{i}-\overline{Z})^{2}\sim\chi^{2}_{(n-1)}$$

MY SOLUTION

(a) Once these random variables are independent, we obtain $$\overline{Z}$$ \begin{align*} M_{\overline{Z}}(t) = \prod_{i=1}^{n} M_{Z_{i}}(t/n) = \prod_{i=1}^{n}\exp\left(\frac{t^{2}}{2n^{2}}\right) = \exp\left(\sum_{i=1}^{n}\frac{t^{2}}{n^{2}}\right) = \exp\left(\frac{t^{2}}{n}\right) \end{align*}

which is the moment generating function of $$\mathcal{N}(0,1/n)$$.

(b) One possible approach consists in determining their joint probability density function and prove that it is the product of its marginals. Another possible approach consists in determining its joint probability density function and use the moment generating functions in order to show they are independent. Besides from these strategies, I do not know how to solve the problem.

Could someone help me with the calculations?

(c) I do not know how to approach.

EDIT

(c) After some thought, I came up with the following solution. To begin, notice that \begin{align*} Y = \sum_{i=1}^{n}(Z_{i} - \overline{Z})^{2} & = \sum_{i=1}^{n}Z^{2}_{i} - 2\sum_{i=1}^{n}Z_{i}\overline{Z} + n\overline{Z}^{2} = \sum_{i=1}^{n}Z^{2}_{i} - 2n\overline{Z}^{2} + n\overline{Z}^{2}\\\\ & = \sum_{i=1}^{n}Z_{i}^{2} - n\overline{Z}^{2} \Rightarrow \sum_{i=1}^{n}Z^{2}_{i} = n\overline{Z}^{2} + \sum_{i=1}^{n}(Z_{i}-\overline{Z})^{2} \end{align*}

If we agree that \begin{align*} U = n\overline{Z}^{2}\quad\text{and}\quad V = \sum_{i=1}^{n}Z^{2}_{i} \end{align*}

since we know that $$Y$$ and $$U$$ are independent, we obtain \begin{align*} M_{V}(t) = M_{U}(t)M_{Y}(t) \Leftrightarrow M_{Y}(t) = \frac{M_{V}(t)}{M_{U}(t)} \end{align*}

given that $$V\sim\chi^{2}_{(n)}$$ and $$U\sim\chi^{2}_{(1)}$$, we conclude that $$Y\sim\chi^{2}_{(n-1)}$$.

• I doubt $(\overline{Z} - Z_{i})^{2}\sim\chi^{2}_{(1)}$ is correct, especially if $n=1$. Indeed, I suspect you should be aiming to prove $\sum\limits_{i=1}^{n}(Z_{i}-\overline{Z})^{2}\sim\chi^{2}_{(n-1)}$. You might want to read about Cochran's_theorem – Henry Feb 6 at 15:41
• Thanks for the comment, Henry. I've written a typo. Now I corrected it. – APC89 Feb 6 at 16:43

for b) try to prove that $$\overline{Z}_n,Z_1-\overline{Z_n},...,Z_n-\overline{Z}_n$$ by using characteristic functions or the properties of multivariate normal distribution...