It is easy to show that when $X_i$ are $iid$ normally distributed with mean $\mu<\infty$ and variance $\sigma^2 >0$.
$$ (n-1)\frac{s^2_{x}}{\sigma^2} \xrightarrow{d} \chi^2_{n-1} $$
But If we assume $X_i$ to be $iid$, but not normally distributed, how do I show that $$ (n-1)\frac{s^2_{x}}{\sigma^2} \xrightarrow{d} \chi^2_{n-1} $$ Or will it not be distributed as chi-sqaure?
For independent normal observations $X_1, \dots, X_n,$ one has $Q = (n-1)S^2/\sigma^2 \sim Chisq(df = n-1),$ so that $E(Q) = n$ and $V(Q) = 2n.$
When data are not normal, it is not true that $Q$ has a chi-squared distribution.
Here is a simulation in R statistical software. The histogram (left panel below) of 100,000 values of $Q$ from random samples if size $n = 5$ from the standard uniform distribution $Unif(0,1)$ clearly does not match the density curve of $Chisq(4).$
By contrast, a similar simulation using standard normal data, gives $E(Q) \approx 4$ and $V(Q) \approx 8,$ within simulation error. And the histogram (right panel) of simulated values is a good fit to the density of $Chisq(4).$
m = 10^5; n = 5; x = runif(m*n) # standard uniform data
DTA = matrix(x, nrow=m) # each row a sample of size 5
Q.u = apply(DTA, 1, var)*(n-1)/(1/12) # m-vector of Q.u values
m = 10^5; n = 5; x = rnorm(m*n) # standard normal data
DTA = matrix(x, nrow=m)
Q.n = apply(DTA, 1, var)*(n-1)
mean(Q.n)
## 4.003542 # consistent with E(Q) = 4
var(Q.n)
## 8.023128 # consistent with V(Q) = 8
Perhaps the following histograms of simulated distributions of $Q$ for samples of size $n = 50$ from standard uniform and of size $n = 500$ from $Exp(rate=1)$ will suggest what happens as $n$ becomes large.
