[proofplan] We first apply the [Central Limit Theorem](/theorems/521) to the standardized sample mean with the true standard deviation $\sigma$. Next we prove that the sample variance $s_n^2$ converges in probability to $\sigma^2$ by rewriting it in terms of empirical first and second moments and applying the [Weak Law of Large Numbers](/theorems/1127). This implies \begin{align*} \frac{s_n}{\sigma}\to 1 \end{align*} in probability, so Slutsky's Theorem gives asymptotic normality of the studentized statistic. Finally, we rewrite the confidence event in terms of that statistic, control the exceptional event $s_n=0$, and evaluate the limiting standard normal probability at the prescribed quantile. [/proofplan] [step:Apply the central limit theorem to the standardized sample mean] Define the standardized sample-mean statistic $X_n:\Omega\to\mathbb R$ by \begin{align*} X_n:=\frac{\sqrt n(\hat\mu_n-\mu)}{\sigma}. \end{align*} Since $Y_1,Y_2,\dots$ are independent identically distributed real-valued random variables with finite mean $\mu$ and finite positive variance $\sigma^2$, the [Central Limit Theorem](/theorems/1848) applies and gives \begin{align*} X_n\xrightarrow{d}Z, \end{align*} where $Z\sim\mathcal N(0,1)$. [/step] [step:Prove that the sample variance converges to the true variance] Define the empirical second moment $M_n:\Omega\to\mathbb R$ by \begin{align*} M_n:=\frac{1}{n}\sum_{i=1}^{n}Y_i^2. \end{align*} Since $\operatorname{Var}(Y_1)=\sigma^2<\infty$ and $\mathbb E[Y_1]=\mu$, we have $\mathbb E[Y_1^2]=\sigma^2+\mu^2<\infty$. The [Weak Law of Large Numbers](/theorems/1851) applied to $Y_1,Y_2,\dots$ gives \begin{align*} \hat\mu_n\xrightarrow{\mathbb P}\mu, \end{align*} and the Weak Law of Large Numbers applied to $Y_1^2,Y_2^2,\dots$ gives \begin{align*} M_n\xrightarrow{\mathbb P}\mathbb E[Y_1^2]=\sigma^2+\mu^2. \end{align*} For each $n\ge 2$, expanding the square in the definition of $s_n^2$ gives \begin{align*} s_n^2=\frac{n}{n-1}\left(M_n-\hat\mu_n^2\right). \end{align*} By the [continuous mapping theorem](/theorems/1847) for multiplication and squaring, together with convergence of the deterministic factor \begin{align*} \frac{n}{n-1}\to 1, \end{align*} it follows that \begin{align*} s_n^2\xrightarrow{\mathbb P}\sigma^2+\mu^2-\mu^2=\sigma^2. \end{align*} [guided] The goal of this step is to replace the unknown variance $\sigma^2$ by the observable sample variance $s_n^2$. To do this, we express $s_n^2$ using empirical moments. Define the empirical second moment $M_n:\Omega\to\mathbb R$ by \begin{align*} M_n:=\frac{1}{n}\sum_{i=1}^{n}Y_i^2. \end{align*} The hypothesis $\operatorname{Var}(Y_1)=\sigma^2<\infty$ and $\mathbb E[Y_1]=\mu$ implies \begin{align*} \mathbb E[Y_1^2]=\operatorname{Var}(Y_1)+(\mathbb E[Y_1])^2=\sigma^2+\mu^2<\infty. \end{align*} Thus both the variables $Y_i$ and the variables $Y_i^2$ satisfy the integrability hypotheses needed for the Weak Law of Large Numbers. Applying the Weak Law of Large Numbers to the independent identically distributed sequence $Y_1,Y_2,\dots$ gives \begin{align*} \hat\mu_n\xrightarrow{\mathbb P}\mu. \end{align*} Applying it again to the independent identically distributed sequence $Y_1^2,Y_2^2,\dots$ gives \begin{align*} M_n\xrightarrow{\mathbb P}\mathbb E[Y_1^2]=\sigma^2+\mu^2. \end{align*} Now we connect these two convergences to $s_n^2$. Expanding the square in the definition of the sample variance, \begin{align*} \sum_{i=1}^{n}(Y_i-\hat\mu_n)^2=\sum_{i=1}^{n}Y_i^2-2\hat\mu_n\sum_{i=1}^{n}Y_i+n\hat\mu_n^2. \end{align*} Since $\sum_{i=1}^{n}Y_i=n\hat\mu_n$, the middle and final terms combine as \begin{align*} -2\hat\mu_n\sum_{i=1}^{n}Y_i+n\hat\mu_n^2=-2n\hat\mu_n^2+n\hat\mu_n^2=-n\hat\mu_n^2. \end{align*} Therefore \begin{align*} \sum_{i=1}^{n}(Y_i-\hat\mu_n)^2=nM_n-n\hat\mu_n^2, \end{align*} and hence \begin{align*} s_n^2=\frac{n}{n-1}\left(M_n-\hat\mu_n^2\right). \end{align*} Because $\hat\mu_n\xrightarrow{\mathbb P}\mu$, the continuous mapping theorem gives $\hat\mu_n^2\xrightarrow{\mathbb P}\mu^2$. Combining this with $M_n\xrightarrow{\mathbb P}\sigma^2+\mu^2$ and the deterministic convergence $n/(n-1)\to 1$, we obtain \begin{align*} s_n^2\xrightarrow{\mathbb P}\sigma^2+\mu^2-\mu^2=\sigma^2. \end{align*} This is the consistency of the sample variance. [/guided] [/step] [step:Use Slutsky's theorem to studentize the statistic] Since $s_n^2\xrightarrow{\mathbb P}\sigma^2$ and $s_n,\sigma\ge 0$ with $\sigma>0$, continuity of the square-root map on $[0,\infty)$ gives \begin{align*} s_n\xrightarrow{\mathbb P}\sigma. \end{align*} Hence \begin{align*} \frac{\sigma}{s_n}\mathbb 1_{\{s_n>0\}}+\mathbb 1_{\{s_n=0\}}\xrightarrow{\mathbb P}1. \end{align*} The reciprocal factor that appears below also converges to $1$ in probability. Indeed, \begin{align*} \left|\frac{\sigma}{s_n}\mathbb 1_{\{s_n>0\}}-\left(\frac{\sigma}{s_n}\mathbb 1_{\{s_n>0\}}+\mathbb 1_{\{s_n=0\}}\right)\right| =\mathbb 1_{\{s_n=0\}}, \end{align*} and $\mathbb P(s_n=0)\to 0$ because $\{s_n=0\}\subset\{|s_n-\sigma|\ge \sigma/2\}$ for every $n$. Therefore \begin{align*} \frac{\sigma}{s_n}\mathbb 1_{\{s_n>0\}}\xrightarrow{\mathbb P}1. \end{align*} Define the modified studentized statistic $T_n:\Omega\to\mathbb R$ by \begin{align*} T_n:=\frac{\sqrt n(\hat\mu_n-\mu)}{s_n}\mathbb 1_{\{s_n>0\}}. \end{align*} On $\{s_n>0\}$, this is the product of $X_n$ and $\sigma/s_n$. Since $X_n\xrightarrow{d}Z$ and the multiplicative factor converges in probability to $1$, Slutsky's Theorem gives \begin{align*} T_n\xrightarrow{d}Z. \end{align*} [/step] [step:Rewrite the confidence event using the studentized statistic] Let $z:=z_{1-\alpha/2}$. Since $\alpha\in(0,1)$, we have $1-\alpha/2\in(1/2,1)$, and therefore $z>0$ for the standard normal distribution. Define the confidence event $A_n\in\mathcal F$ by \begin{align*} A_n:=\left\{\mu\in\left[\hat\mu_n-z\frac{s_n}{\sqrt n},\hat\mu_n+z\frac{s_n}{\sqrt n}\right]\right\}. \end{align*} On the event $\{s_n>0\}$, the event $A_n$ is equivalent to \begin{align*} |T_n|\le z. \end{align*} Indeed, subtracting $\hat\mu_n$ from the interval inequalities, multiplying by $\sqrt n/s_n>0$, and taking absolute values gives exactly this condition. Therefore \begin{align*} \left|\mathbb P(A_n)-\mathbb P(|T_n|\le z)\right|\le \mathbb P(s_n=0). \end{align*} Because $s_n\xrightarrow{\mathbb P}\sigma$ and $\sigma>0$, the inclusion \begin{align*} \{s_n=0\}\subset\{|s_n-\sigma|\ge \sigma/2\} \end{align*} implies \begin{align*} \mathbb P(s_n=0)\to 0. \end{align*} Thus $A_n$ has the same limiting probability as the event $\{|T_n|\le z\}$. [/step] [step:Evaluate the limiting normal probability at the quantile] Since $T_n\xrightarrow{d}Z$ and the standard normal distribution has no atoms at either $z$ or $-z$, convergence in distribution gives \begin{align*} \mathbb P(|T_n|\le z)\to \mathbb P(|Z|\le z). \end{align*} Using the symmetry of the standard normal distribution and the defining property $\mathbb P(Z\le z)=1-\alpha/2$, we compute \begin{align*} \mathbb P(|Z|\le z)=\mathbb P(-z\le Z\le z)=\mathbb P(Z\le z)-\mathbb P(Z<-z). \end{align*} By symmetry, $\mathbb P(Z<-z)=\mathbb P(Z>z)=\alpha/2$, so \begin{align*} \mathbb P(|Z|\le z)=1-\alpha/2-\alpha/2=1-\alpha. \end{align*} Combining this limit with the estimate from the preceding step yields \begin{align*} \mathbb P\left(\mu\in\left[\hat\mu_n-z_{1-\alpha/2}\frac{s_n}{\sqrt n},\hat\mu_n+z_{1-\alpha/2}\frac{s_n}{\sqrt n}\right]\right)\to1-\alpha. \end{align*} This is the desired asymptotic coverage statement. [/step]