# How to derive the asymptotic distribution of t-statistic?

Let $${X_n}$$ be an IID sample such that $${X_i} \sim N(\mu,\sigma^2)$$. When both $$\mu$$ and $$\sigma$$ are unknown, we construct $$t(\hat{\mu},s)=\dfrac{\sqrt{n}(\hat{\mu}-\mu)}{s}$$, where $$s$$ is the sample standard deviation.

The statistic $$t(\hat{\mu},s)$$ follows t-distribution exactly. I wish to know the asymptotic distribution of $$t(\hat{\mu},s)$$.

I understand that by Continuous Mapping theorem under certain conditions, if $$\hat{\mu} \xrightarrow{d} X \sim N(\mu,\sigma^2/n)$$ then $$h(\hat{\mu}) \xrightarrow{d} h(X)$$. However, in case of $$t(\hat{\mu},s)$$, it is a function of two random variables. So how to derive its asymptotic distribution?

I also checked Slutsky's theorem. That also requires that at least one of $$\hat{\mu}$$ and $$s$$ should converge to a constant. Now by CLT, both converges in distribution to a random variable with normal distribution. However, by combining LLN we may say that $$s \xrightarrow{p} \sigma$$. Would it be the right way to go?

EDIT: As pointed out in the second comment this is not a duplicate of this. Seemingly $$s$$ converged in probability to a constant but to a random variable in distribution. This was the main doubt that whether it was okay to ignore the convergence in distribution.

• – StubbornAtom Mar 30 at 19:26
• @StubbornAtom : I don't agree that this is a duplicate, since this question asks where the gaps are in a proposed argument, rather than just asking what the argument is. – Michael Hardy Mar 30 at 20:09
• What is the difference between $\hat \sigma$ and $s$? – Alecos Papadopoulos Apr 1 at 14:35
• No difference. Used $\hat{\sigma}$ by mistake. Corrected now. – Dayne Apr 2 at 0:52

It is not correct that $$\widehat\mu$$ and $$s$$ fail to converge in distribution to constants, nor that CLT implies that.
\begin{align} & \frac{\widehat\mu - \mu}{\sigma/\sqrt n} \overset d \longrightarrow \operatorname N(0,1) \\[8pt] \text{and } & \sqrt{n-1} \left. \left(\dfrac{s^2}{\sigma^2}-1\right) \right/\!\!\sqrt 2 \overset d \longrightarrow \operatorname N(0,1). \end{align}
It can be shown to follow that $$\widehat\mu$$ and $$s$$ converge in distribution to $$\mu$$ and $$\sigma$$ respectively.