This is my first question here, so excuse the poor formatting. I am working on a statistics assignment and they asked me the following question:
$Xi i.i.d. ∼ N(µ, σ^2)$ Give a 100(1 − α)% two-sided equal-tailed approximate confidence interval for µ based on a limiting distribution of the MLE of µ for a given α and random sample X1,...,Xn.
I have tried two approaches, and I am not sure which one is correct. Can anyone comment on this?
1)
We have the following $(X ̅-μ)/(S⁄√n)$ ~ t(n-1) as n gets large, the t-distribution will come closer and closer to the standard normal distribution. Hence, we can conclude that the limiting distribution of
$ (X ̅-μ)/(S⁄√n)$ is N(0,1). So $(X ̅-μ)/(S⁄√n)→N(0,1)$ as n→∞
2) Using limit distributions, we can say that $S^2→ σ^2$ as n→ ∞. Therefore, we can use the CLT and state the following limiting distribution: $√n (X ̅-μ)→N(0,σ^2)$ as n→ ∞. This means that $(X ̅-μ)/(S_n⁄√n)→N(0,1) as n→ ∞$
From both approaches onwards, I would continue by constructing the simple 100(1-alpha)% confidence interval based on the normal distribution. I am just not sure if either of my approaches to justify normality for large n's is correct.
Thanks a lot in advance!
Use $\bar X \pm t^*S/\sqrt{n},$ where $t^*$ cuts probability $\alpha/2$ from the upper tail of Student's t distribution with $n-1$ degrees of freedom.
Your approach in (1) is headed in the right direction: The MLE of $\mu$ is $\hat\mu = \bar X$ and the MLE of $\sigma^2$ is $\hat{\sigma^2} = \frac{n-1}{n}S^2$ where $S^2$ is the usual sample standard deviation. Although $E(S)$ is not exactly $\sigma,$ the bias disappears as $n \rightarrow \infty.$ Also, as $n \rightarrow \infty,$ we have $T = \sqrt{n}(\bar X - \mu)/S$ converges in distribution to standard normal. (For $n > 30,$ we have $t^* \approx 1.96$ for a 95% CI.)