standard normal distribution with slutsky

Question

i have a small question. We Have $S_{n}$ with $ S_{n}\overset{D}{\longrightarrow}N(0,\sigma^{2})$ (normal distribution). We have also an estimator for $\sigma$, it is $\hat\sigma$. Than we $\hat\sigma^{2}\overset{P}{\longrightarrow}\sigma^{2}$. So my question is: why its follows with Slutksy : $$ \dfrac{S_{n}}{\hat{\sigma^{2}}}\longrightarrow N(0,1)$$

It would be very helpfull, when someone has some tips!

Thank you, greets, Hendrik.

Answer 1

$S_n{\stackrel d\to} N(0,\sigma^2)$ is an abbreviation for $S_n{\stackrel d\to} Z$ where $Z\sim N(0,\sigma^2)$. This might seem pedantic, and in fact ordinarily I wouldn't bother with this point, but you seem to be confused about what $\alpha\cdot N(0,\sigma^2)$ means if $\alpha$ is a constant. Slutsky's theorem tells us that, since $\hat\sigma_n{\stackrel p\to}\sigma$ in probability, $\frac{S_n}{\hat\sigma_n}{\stackrel d\to}\frac{Z}{\sigma}$. If $Z\sim N(0,\sigma^2)$, what is the distribution of $\frac{Z}{\sigma}$?

As a minor point, you should not be looking at $\frac{S_n}{\hat\sigma_n^2}$ (remove the square).

Answer 2

We will apply the following case of Slutsky's Theorem: Let $\{X_n\}$, $\{Y_n\}$ be sequences of random elements. If $X_n$ converges in distribution to a random element $X$ (as $n \rightarrow \infty$) and $Y_n$ converges in probability to a constant $c$ then $X_n Y_n$ converges in distribution to $cX$.

Your assumptions can be stated as

1. ${S_n}/{\sigma}$ converges in distribution to $\mathcal{N}(0,1)$.
2. ${\sigma}/{\hat{\sigma}}$ converges in probability to the number $c = 1$.

Write $$ \frac{S_n}{\hat{\sigma}} = \frac{S_n}{\sigma} \cdot \frac{\sigma}{\hat{\sigma}} $$ Then the result follows immediately from putting $X_n = S_n / \sigma$, $Y_n = \sigma / \hat{\sigma}$.