10 replaced http://stats.stackexchange.com/ with https://stats.stackexchange.com/ edited Apr 13 '17 at 12:44 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 edited Dec 7 '16 at 13:41 Alecos Papadopoulos 43.2k298201 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 edited Nov 17 '14 at 1:10 Alecos Papadopoulos 43.2k298201 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. 7 added 52 characters in body edited Jul 28 '14 at 18:46 Alecos Papadopoulos 43.2k298201 6 deleted 2028 characters in body edited Jul 28 '14 at 18:34 Alecos Papadopoulos 43.2k298201 5 deleted 726 characters in body edited Jul 7 '14 at 19:33 Alecos Papadopoulos 43.2k298201 4 Fixed some convergence arrows and some upper/lower-cases. edited Jul 1 '14 at 19:31 COOLSerdash 16.9k75395 3 Added result edited Jul 1 '14 at 17:43 Alecos Papadopoulos 43.2k298201 2 Added dichotomous variable case edited Jul 1 '14 at 2:43 Alecos Papadopoulos 43.2k298201 1 answered Jul 1 '14 at 0:38 Alecos Papadopoulos 43.2k298201