Why is $\frac{\sum^n_{i=1}(X_i-\bar{X})^2}{\sigma^2}$chi-square distributed with $n-1$ degrees of freedom? [duplicate]

Question

On Wikipedia, it says

If $X_{i};i=1,\ldots ,n$ are independent normal $(\mu ,\sigma ^{2})$ random variables, the statistic

$$\frac{\sum\limits^n_{i=1}(X_i-\bar{X})^2}{\sigma^2}$$

follows a chi-squared distribution with $n − 1$ degrees of freedom. Here, the degrees of freedom arises from the residual sum-of-squares in the numerator, and in turn the $n − 1$ degrees of freedom of the underlying residual vector $\{X_{i}-{\bar {X}}\}$.

With that in mind, I'd like to put forward the following example:

Suppose we choose an integer, $10$. Let's say our goal is to choose some $n$ values that sum to $10$. Let's say $n=3$. I freely choose $6$ and then $3$, but now I am forced to choose $1$. Hence I have $2$ degrees of freedom. Or more generally, $n-1$ degrees of freedom.

In this example, it's clear that the final choice is not a free one, because we have some target value, and so once we've made $n-1$ choices for the summands, there is only one possible value that ensures we satisfy the target sum value constraint.

Now, going back to case of the residual sum-of-squares on the numerator. If this were $\sum\limits^n_{i=1}(X_i-\bar{X})$ instead (notice I've dropped the square), then our target is $0$. So again, like in the example above, we have $n-1$ degrees of freedom because we can make free choices for $X_1\ldots X_{n-1}$, but then $X_n$ is forced. However, once we square each $(X_i - \bar{X})$, we lose the target value of $0$.

So how do we know that the degrees of freedom is $n-1$ without a target value? If there is no target value, why are we not able to make $n$ free choices?

Answer 1

Consider samples of size $n= 5$ from a standard normal distribution. then $Q =(n-1)S^2 \sim \mathsf{Chisq}(\nu=4),$ not $\mathsf{Chisq}(\nu=5).$

set.seed(706);  m = 10^6;  n = 5
q = replicate( m, (n-1)*var(rnorm(n)) )
mean(q)
[1] 4.002257  # aprx E(Q) = 4
hdr = "Simulated Dist'n of Q fits CHISQ(4)[blue], not CHISQ(5) [red]"
hist(q, prob=T, br=50, col="skyblue2", main=hdr)
  curve(dchisq(x, 4), add=T, lwd=2, col="blue")
  curve(dchisq(x, 5), add=T, lwd=3, col="red", lty="dotted")

Note: If you consider $n = 2,$ then it is easy to verify that $\bar X$ is a function of $X_1 + X_2$ and independently, $S^2$ is a function of $X_1 - X_2$ so that $S^2 \sim \mathsf{Chisq}(1).$ One proof for $n > 2$ shows that $\bar X$ is a function of a vector in one dimention, and that (orthogonally) $S^2$ is a function of vectors in $n-1$ dimensions.