The confidence interval for the mean of a random variable $Y$ has coverage $1-\alpha$ which I am trying to show. Starting from $$\widehat{E(Y)} - q_{1-\frac{\alpha}{2}}\sqrt{\frac{\widehat{Var(Y)}}{n}} \leq E(Y) \leq \widehat{E(Y)} + q_{1-\frac{\alpha}{2}}\sqrt{\frac{\widehat{Var(Y)}}{n}} $$ I re-arranged this to $$- q_{1-\frac{\alpha}{2}} \leq \sqrt{n} \left( \frac{\widehat{E(Y)}-E(Y)}{\sqrt{Var(Y)}} \right) \leq q_{1-\frac{\alpha}{2}} $$ Then by the central limit theorem $\sqrt{n} (\widehat{E(Y)} - E(Y)) \rightarrow N(0,Var(Y))$. Then I use the continuous mapping theorem. Given that $\widehat{Var(Y)} = \frac{1}{n}\sum^{n}_{i=1} Y_i^2 - (\frac{1}{n}\sum^n_i Y_i)^2$, define the function $m:(x,y) \rightarrow x - y^2$. Then $$x \equiv \sum^{n}_{i=1} Y_i^2 \rightarrow E(Y^2)$$ and $$y \equiv \sum^{n}_{i=1} Y_i \rightarrow E(Y)$$ by the weak law of large numbers. Then $\widehat{Var(Y)} = g(x,y) \rightarrow g(E(Y^2),E(Y)) = Var(Y)$ is a consistent estimator of the variance. Hence $$\sqrt{n} \left( \frac{\widehat{E(Y)}-E(Y)}{\sqrt{Var(Y)}} \right) \rightarrow N(0,1)$$

But now I am stuck in how to proceed to obtain the final coverage rate. If someone could give me a hint on how to get there, this would be greatly appreciated.


Thanks for pointing out the typo. I think I understand the hint. This should be $$ \begin{align} Pr \left(\sqrt{n} \left( \frac{\widehat{E(Y)}-E(Y)}{\sqrt{\widehat{Var(Y)}}} \in (-q_{1-\frac{\alpha}{2}}, q_{1-\frac{\alpha}{2}} \right) \right) &= Pr(N(0,1) \in (-q_{1-\frac{\alpha}{2}}, q_{1-\frac{\alpha}{2}} )) \newline &= Pr(N(0,1) \leq q_{1-\frac{\alpha}{2}}) - Pr(N(0,1) > q_{\frac{\alpha}{2}}) \newline &= \Phi (q_{1- \frac{\alpha}{2}}) - \Phi (q_{\frac{\alpha}{2}}) \newline &= 1 - \frac{\alpha}{2} - \frac{\alpha}{2} \newline &= 1 - \alpha \end{align} $$


Probably because you got the quantiles wrong (I suppose your $q(\cdot )$ are the quantiles of the standard normal), which should be $q_{1-\frac{\alpha}{2}}$. Basically what you want to show is that $$Pr\left( E(Y) \in \left( \widehat{E(Y)} - q_{1-\frac{\alpha}{2}} \sqrt{\frac{\widehat{Var(Y)}}{n}}, \widehat{E(Y)} + q_{1-\frac{\alpha}{2}} \sqrt{\frac{\widehat{Var(Y)}}{n}} \right) \right) $$ which is equal to showing that $$Pr\left( \sqrt{n}\left( \frac{\widehat{E(Y)}-E(Y)}{\sqrt{\widehat{Var(Y)}}} \right) \in \left( -q_{1-\frac{\alpha}{2}}, q_{1-\frac{\alpha}{2}} \right) \right) $$ You already know that $\sqrt{n}\left( \frac{\widehat{E(Y)}-E(Y)}{\sqrt{\widehat{Var(Y)}}} \right)$ is distributed as $N(0,1)$ so this should be straight forward. A trick that will help is to know that $-q_{1-\frac{\alpha}{2}}=q_{\frac{\alpha}{2}}$ in order to figure out the probabilities that $N(0,1) \leq q_{1-\frac{\alpha}{2}}$ and $N(0,1) > q_\frac{\alpha}{2}$

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