Otdfbt

In the notes that I'm working through it says the following:
"Let $X_1,...,X_n$ be a random sample from $N(mu,sigma)$

$$sum^n_i=1Bigg[frac(X_i-mu)sigmaBigg]^2$$ has a $chi^2$ distribution with $n$ degrees of freedom.

Now if we modify this by replacing $mu$ with $overlineX$ the distribution changes and we obtain:
$$sum^n_i=1Bigg[frac(X_i-overlineX)sigmaBigg]^2$$ has a $chi^2$ distribution with $n-1$ degrees of freedom."

My question is: why does the number of degrees of freedom change?

My understanding of what a degree of freedom is that the number of degrees of freedom is the number of values in the final calculation of a statistic that are free to vary. So surely as there are $n$ $X_i$ even when we introduce $overlineX$ the number of values that are free to change in the calculation of the statistic is still the same??

Suppose we have a random sample from $mathsfNorm(mu, sigma).$

Let the $V_1 = S^2 = frac1n-1sum_i=1^n (X_i - bar X)^2$ be the estimate of the population variance $sigma^2$ when $mu$ is unknown and estimated by $bar X.$ Then $Q_1 = frac(n-1)V_1sigma^2 sim mathsfChisq(n-1).$

Let the $V_2 = frac1nsum_i=1^n (X_i - mu)^2$ be the estimate of the population variance $sigma^2$ when $mu$ is known. Then $Q_2 = fracnV_2sigma^2 sim mathsfChisq(n).$

In the R code below, we choose $m = 10^6$ samples of size $n = 5$ from $mathsfNorm(mu = 50, sigma = 7).$
Then we make histograms of $Q_1$ and $Q_2.$ Chi-squared densities with degrees of freedom $4$ and $5,$ respectively, fit the histograms. The density curve
for $mathsfChisq(5)$ does not fit the histogram
for $Q_1.$

set.seed(2019)
m = 10^6; n = 5; mu = 50; sg = 7
x = rnorm(m*n, mu, sg)
MAT = matrix(x, nrow=m) # each row a sample of n
v1 = apply(MAT, 1, var) # uses sample mean
v2 = rowSums((MAT - mu)^2)/n # uses population mean
q1 = (n-1)*v1 /sg^2
q2 = n*v2 / sg^2 

mean(q1); var(q1)
[1] 3.997226 # aprx E(Q1) = 4
[1] 8.00637 # aprx Var(Q1) = 8
mean(q2); var(q2)
[1] 4.997005 # aprx E(Q2) = 5
[1] 9.98925 # aprx Var(Q2) = 10

par(mfrow=c(1,2)) # enables 2 panels per plot
hist(q1, prob=T, br=50, col="skyblue2", ylim=c(0,.2))
 curve(dchisq(x, n-1), add=T, lwd=2)
 curve(dchisq(x, n), add=T, col="red", lwd=2, lty="dotted")
hist(q2, prob=T, br=50, col="skyblue2")
 curve(dchisq(x, n), add=T, lwd=2) 
par(mfrow=c(1,1)) 

Your understanding of degrees of freedom is correct for one of the two senses of the term. The vector $left(X_1-overline X,ldots,X_n-overline Xright)$ has $n-1$ degrees of freedom because it is subject to the constraint that the sum of the components must be $0,$ so if you know $n-1$ of them plus that constraint, then you know all of them.

The other sense of the term degrees of freedom is used when one speaks of a chi-square distribution with a specified number of degrees of freedom. Consider any orthonormal basis of the $(n-1)$-dimensional space of $n$-tuples in which the sum of the components is $0.$ Let $U_1,ldots,U_n-1$ be the scalar components of $left( X_1-overline X, ldots, X_n-overline X right)$ with respect to that basis. Then
beginalign
& left( X_1-overline Xright)^2 + cdots + left( X_n - overline Xright)^2 = U_1^2 + cdots + U_n-1^2 \[5pt]
& textand U_1,ldots,U_n-1 sim texti.i.d. N(0, sigma^2).
endalign

Your understanding of degrees of freedom is correct. The difference essentially boils down to a subtle difference between $mu$ and $barX$.

The sample mean $barX$ is determined by the values of the observed samples. As a result, there's some redundancy between $barX$ and the individual values of $x_i$. Suppose we knew $barX$ and the first $N-1$ values. That's enough, because we can write out the equation for $barX$ as
$$barX=frac1Nx_1 +frac1Nx_2 + frac1Nx_3+ ldots + frac1Nx_N-1 + frac1Nx_N$$

A little bit of rearrangement lets us find that missing value: $$x_N = barX - fracN-1N(x_1+x_2+ldots +x_N-1) $$

This isn't the case when the population mean ($mu$) is used, since it is not dependent on the observed data; it's known (or assumed to be known) ahead of time. You therefore need to use all $N$ values to produce calculate your test statistic.