# Showing that the sample mean and variance are independent by showing two distribution are the same

Let $$X_{1},X_{2},....,X_{n}$$ be independent $$N(0,1)$$ random variables and set $$\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}$$ show that $$\bigg(\bar{X},\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\bigg)$$ and $$\bigg({X}/\sqrt{n},\frac{1}{n}\sum_{i=1}^{n-1}(X_{i})^{2}\bigg)$$ have the same distribution. This shows that the sample mean and sample variance are independent in this model.

Thoughts I tried showing that they have the same characteristic function but couldn't get it to work that way. I think maybe I could do something involving the CLT but I'm not sure how.

• – StubbornAtom Feb 13 at 18:44

Using the Gram-Schmidt process, construct an orthonormal basis of $${\mathbb R}^n$$ containing the unit vector $${\bf a}:=\frac1{\sqrt n}(1,\ldots,1)^T$$. Convert this basis into an orthogonal matrix $$A$$ with $$\bf a$$ as its last row. That is, we have found an $$n\times n$$ matrix $$A$$ with $$AA^T=A^TA=I$$ and $$A_{ni}=\frac1{\sqrt n}$$ for each $$i$$.

Pack the given variables $$X_1,\ldots,X_n$$ into a column vector $$X$$ and define a new vector of random variables $$Z:=(Z_1,\ldots,Z_n)^T$$ by $$Z=AX$$. We observe:

1. The variables $$Z_1,\ldots,Z_n$$ are also independent standard normal variables, since they are a linear combination of the $$X$$'s, with mean vector zero and covariance matrix $$\operatorname{Cov}(Z)=E(ZZ^T)=E(AXX^TA^T)=AE(XX^T)A^T=AA^T=I.$$
2. The final variable $$Z_n$$ equals $$\sqrt n\bar X$$, since $$Z_n=(AX)_n=\sum_iA_{ni}X_i=\sum\frac1{\sqrt n} X_i=\frac1{\sqrt n}n\bar X.$$
3. The sum $$\sum_{i=1}^nZ_i^2$$ equals $$\sum_{i=1}^nX_i^2$$, since $$\sum_{i=1}^nZ_i^2=Z^TZ=(AX)^T(AX)=X^TA^TAX=X^TX=\sum_{i=1}^nX_i^2.$$
4. The variable $$\sum_{i=1}^{n-1}Z_i^2$$ equals $$\sum_{i=1}^nZ_i^2-Z_n^2=\sum_{i=1}^nX_i^2-n(\bar X)^2$$ by (2) and (3). This last can be rewritten as $$\sum_{i=1}^n(X_i-\bar X)^2$$.

We conclude that the pair $$(\sqrt n\bar X,\sum_{i=1}^n(X_i-\bar X)^2)$$ is identical to the pair $$(Z_n,\sum_{i=1}^{n-1}Z_i^2)$$, which by (1) has the same distribution as $$(X_n,\sum_{i=1}^{n-1}X_i^2)$$.

This result can be generalized: Let $$X_1,\ldots,X_n$$ be iid, each with standard normal distribution. If $$\alpha_1,\ldots,\alpha_n$$ are constants with $$\sum\alpha_i^2=1$$, then $$\sum_{i=1}^nX_i^2-\left(\sum_{i=1}^n\alpha_iX_i\right)^2$$ has chi-squared distribution with $$n-1$$ degrees of freedom and is independent of $$\sum\alpha_iX_i$$, which has standard normal distribution.