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To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$. The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,

$$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answerthis answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$. The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,

$$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$. The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,

$$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

9 Shortened and clarified the proof
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To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.  

From the remaining terms, let's first examine the termWe also have $W_n \equiv \sqrt n\Big(\bar x-\mu\Big)^2$$\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$.
We know that $$Z_n = \sqrt n\frac {\bar x-\mu}{\sigma} \xrightarrow{d} Z \sim N(0,1)$$

Applying The first component converges in distribution to a Normal, the continuous mappingsecond convergres in probability to zero. Then by Slutsky's theorem we have that

$$Q_n = n\left(\frac {\bar x-\mu}{\sigma}\right)^2=Z_n^2 \xrightarrow{d} Z^2 \sim \mathcal \chi^2_1$$

Therefore

$$G_n = \sigma^2Q_n = n\Big(\bar x-\mu\Big)^2 \xrightarrow{d} \sigma^2Z^2 \sim \mathrm{Gamma}(k=1/2, \theta = 2\sigma^2)$$

We see that the term of interest can be written

$$W_n = \frac 1{\sqrt n}G_n$$

Consider whether itsproduct converges in probability limit isto zero: $$\lim_{n \rightarrow \infty}P(|W_n|>\epsilon) = \lim_{n \rightarrow \infty}P\left(\left|\frac 1{\sqrt n}G_n\right|>\epsilon\right) = \lim_{n \rightarrow \infty}P\left(G_n>\sqrt n\epsilon\right) = P\left(\sigma^2Z^2>\epsilon\lim_{n \rightarrow \infty}\sqrt n\right)=0$$,

since the right-hand side of the inequality inside the probability goes to infinity, and $\sigma^2Z^2$ has a well-defined distribution. So $$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.  

From the remaining terms, let's first examine the term $W_n \equiv \sqrt n\Big(\bar x-\mu\Big)^2$.
We know that $$Z_n = \sqrt n\frac {\bar x-\mu}{\sigma} \xrightarrow{d} Z \sim N(0,1)$$

Applying the continuous mapping theorem we have that

$$Q_n = n\left(\frac {\bar x-\mu}{\sigma}\right)^2=Z_n^2 \xrightarrow{d} Z^2 \sim \mathcal \chi^2_1$$

Therefore

$$G_n = \sigma^2Q_n = n\Big(\bar x-\mu\Big)^2 \xrightarrow{d} \sigma^2Z^2 \sim \mathrm{Gamma}(k=1/2, \theta = 2\sigma^2)$$

We see that the term of interest can be written

$$W_n = \frac 1{\sqrt n}G_n$$

Consider whether its probability limit is zero: $$\lim_{n \rightarrow \infty}P(|W_n|>\epsilon) = \lim_{n \rightarrow \infty}P\left(\left|\frac 1{\sqrt n}G_n\right|>\epsilon\right) = \lim_{n \rightarrow \infty}P\left(G_n>\sqrt n\epsilon\right) = P\left(\sigma^2Z^2>\epsilon\lim_{n \rightarrow \infty}\sqrt n\right)=0$$

since the right-hand side of the inequality inside the probability goes to infinity, and $\sigma^2Z^2$ has a well-defined distribution. So $$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$. The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,

$$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

8 added 172 characters in body
source | link

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

From the remaining terms, let's first examine the term $W_n \equiv \sqrt n\Big(\bar x-\mu\Big)^2$.
We know that $$Z_n = \sqrt n\frac {\bar x-\mu}{\sigma} \xrightarrow{d} Z \sim N(0,1)$$

Applying the continuous mapping theorem we have that

$$Q_n = n\left(\frac {\bar x-\mu}{\sigma}\right)^2=Z_n^2 \xrightarrow{d} Z^2 \sim \mathcal \chi^2_1$$

Therefore

$$G_n = \sigma^2Q_n = n\Big(\bar x-\mu\Big)^2 \xrightarrow{d} \sigma^2Z^2 \sim \mathrm{Gamma}(k=1/2, \theta = 2\sigma^2)$$

We see that the term of interest can be written

$$W_n = \frac 1{\sqrt n}G_n$$

Consider whether its probability limit is zero: $$\lim_{n \rightarrow \infty}P(|W_n|>\epsilon) = \lim_{n \rightarrow \infty}P\left(\left|\frac 1{\sqrt n}G_n\right|>\epsilon\right) = \lim_{n \rightarrow \infty}P\left(G_n>\sqrt n\epsilon\right) = P\left(\sigma^2Z^2>\epsilon\lim_{n \rightarrow \infty}\sqrt n\right)=0$$

since the right-hand side of the inequality inside the probability goes to infinity, and $\sigma^2Z^2$ has a well-defined distribution. So $$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

From the remaining terms, let's first examine the term $W_n \equiv \sqrt n\Big(\bar x-\mu\Big)^2$.
We know that $$Z_n = \sqrt n\frac {\bar x-\mu}{\sigma} \xrightarrow{d} Z \sim N(0,1)$$

Applying the continuous mapping theorem we have that

$$Q_n = n\left(\frac {\bar x-\mu}{\sigma}\right)^2=Z_n^2 \xrightarrow{d} Z^2 \sim \mathcal \chi^2_1$$

Therefore

$$G_n = \sigma^2Q_n = n\Big(\bar x-\mu\Big)^2 \xrightarrow{d} \sigma^2Z^2 \sim \mathrm{Gamma}(k=1/2, \theta = 2\sigma^2)$$

We see that the term of interest can be written

$$W_n = \frac 1{\sqrt n}G_n$$

Consider whether its probability limit is zero: $$\lim_{n \rightarrow \infty}P(|W_n|>\epsilon) = \lim_{n \rightarrow \infty}P\left(\left|\frac 1{\sqrt n}G_n\right|>\epsilon\right) = \lim_{n \rightarrow \infty}P\left(G_n>\sqrt n\epsilon\right) = P\left(\sigma^2Z^2>\epsilon\lim_{n \rightarrow \infty}\sqrt n\right)=0$$

since the right-hand side of the inequality inside the probability goes to infinity, and $\sigma^2Z^2$ has a well-defined distribution. So $$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

To side-step dependencies arising when we consider the sample variance, we write

$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$

and after a little manipualtion,

$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$

Therefore

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$

Manipulating,

$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$

$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$

The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.

From the remaining terms, let's first examine the term $W_n \equiv \sqrt n\Big(\bar x-\mu\Big)^2$.
We know that $$Z_n = \sqrt n\frac {\bar x-\mu}{\sigma} \xrightarrow{d} Z \sim N(0,1)$$

Applying the continuous mapping theorem we have that

$$Q_n = n\left(\frac {\bar x-\mu}{\sigma}\right)^2=Z_n^2 \xrightarrow{d} Z^2 \sim \mathcal \chi^2_1$$

Therefore

$$G_n = \sigma^2Q_n = n\Big(\bar x-\mu\Big)^2 \xrightarrow{d} \sigma^2Z^2 \sim \mathrm{Gamma}(k=1/2, \theta = 2\sigma^2)$$

We see that the term of interest can be written

$$W_n = \frac 1{\sqrt n}G_n$$

Consider whether its probability limit is zero: $$\lim_{n \rightarrow \infty}P(|W_n|>\epsilon) = \lim_{n \rightarrow \infty}P\left(\left|\frac 1{\sqrt n}G_n\right|>\epsilon\right) = \lim_{n \rightarrow \infty}P\left(G_n>\sqrt n\epsilon\right) = P\left(\sigma^2Z^2>\epsilon\lim_{n \rightarrow \infty}\sqrt n\right)=0$$

since the right-hand side of the inequality inside the probability goes to infinity, and $\sigma^2Z^2$ has a well-defined distribution. So $$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$

We are left with the term

$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$

Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have

$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$

and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and

$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$

Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.

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