Photo of Evgeny Evgenievich Slutsky. Sources: MacTutor and Bomkj.

Applying the continuous mapping theorem and $latex {(v)}&fg=000000$ from the last post, we get the following theorem

Lemma (Slutsky). Let be $latex {X_{n}}&fg=000000$, $latex {X}&fg=000000$ and $latex {Y_{n}}&fg=000000$ random vectors and $latex {c}&fg=000000$ a constant vector. If $latex {X_{n}\rightsquigarrow X}&fg=000000$ and $latex {Y_{n}\rightsquigarrow c}&fg=000000$, then

$latex {(i)}&fg=000000$ $latex {X_{n}+Y_{n}\rightsquigarrow X+c}&fg=000000$.
$latex {(ii)}&fg=000000$ $latex {Y_{n}X_{n}\rightsquigarrow cX}&fg=000000$.
$latex {(iii)}&fg=000000$ $latex {Y_{n}^{-1}X_{n}\rightsquigarrow c^{-1}X}&fg=000000$ provided $latex {c\neq0}&fg=000000$.

The constant in $latex {(i)}&fg=000000$ should be a vector of the same dimension of $latex {X}&fg=000000$. For $latex {(ii)}&fg=000000$ and $latex {(iii)}&fg=000000$ it is also true for $latex {Y_{n}}&fg=000000$ and $latex {c}&fg=000000$ constants and matrices, provided that $latex {c\neq0}&fg=000000$ means $latex {c}&fg=000000$ being invertible.

Plot of the cumulative
Plot of the cumulative distribution functions of several members of the Student t-family of probability distributions.

Example ($latex {t}&fg=000000$-statistics) Let $latex {Y_{1},Y_{2},\ldots}&fg=000000$ be independent, identically, distributed random variables, with $latex {\mathbb E\mbox{\ensuremath{\left(Y_{1}\right)=0}}}&fg=000000$and $latex {\mathbb E\left(Y_{2}^{2}\right)<\infty}&fg=000000$. We pose the question: What is the $latex {\sqrt{n}\overline{Y}_{n}/S_{n}}&fg=000000$ distribution, where $latex {S_{n}^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(Y_{i}-\overline{Y}_{n})^{2}}&fg=000000$ is the sample variance?

Foremost, applying twice the weak law of large numbers $latex {\frac{1}{n}\sum_{i=1}^{n}Y_{i}^{2}}&fg=000000$ and $latex {\overline{Y}_{n}}&fg=000000$, and using the continuous mapping theorem for convergence in probability (with $latex {g(x,y)=x-y^{2}}&fg=000000$), we note that

$latex \displaystyle S_{n}^{2}=\frac{n}{n-1}\left(\frac{1}{n}\sum_{i=1}^{n}Y_{i}^{2}-\overline{Y}_{n}^{2}\right)\stackrel{P}{\rightarrow}1\left(\mathbb E Y_{1}^{2}-\left(EY_{1}\right)^{2}\right)=\mathop{\mathrm{Var}}\left(Y_{1}\right). &fg=000000$

Again by the continuous mapping theorem, $latex {S_{n}}&fg=000000$ converges in probability to $latex {\mathrm{Std}\left(Y_{1}\right)}&fg=000000$. By the central limit theorem, $latex {\sqrt{n}\overline{Y}_{n}}&fg=000000$ converges in law to a $latex {N\left(0,\mathop{\mathrm{Var}}\left(Y_{1}\right)\right)}&fg=000000$. Finally, Slutsky’s lemma gives that $latex {t}&fg=000000$-statistics $latex {\sqrt{n}\overline{Y}_{n}/S_{n}}&fg=000000$ converge to

$latex \displaystyle N\left(0,\mathop{\mathrm{Var}}\left(Y_{1}\right)\right)/\mathrm{Std}\left(Y_{1}\right)=N(0,1). &fg=000000$

If the distribution function of $latex X &fg=000000$ is continuous then $latex {X_{n}\rightsquigarrow X}&fg=000000$  implies that $latex \mathbb P\left(X_{n}\leq x\right)-\mathbb P\left(X\leq x\right) &fg=000000$ for every $latex x &fg=000000$ uniformly.

Lemma. We suppose that $latex {X_{n}\rightsquigarrow X}&fg=000000$ and the distribution function of $latex {X}&fg=000000$ is continuous. Then

$latex \displaystyle \sup_{x}|\mathbb P\left(X_{n}\leq x\right)-\mathbb P\left(X\leq x\right)|\rightarrow0. &fg=000000$

Proof: We will show it only for dimension one, the same idea applies to higher dimensions. We denote $latex {F}&fg=000000$ and $latex {F_{n}}&fg=000000$ the distribution functions of $latex {X}&fg=000000$ and $latex {X_{n}}&fg=000000$. Let $latex {\epsilon>0}&fg=000000$ and $latex {k}&fg=000000$ an integer such that $latex {1/k\leq\epsilon}&fg=000000$. Since $latex {F}&fg=000000$ is bounded and continuous , there are $latex x_{1}, \ldots ,x_{k}&fg=000000$ such $latex {F(x_{i})=i/k}&fg=000000$. Let $latex {x_{i-1}\leq x\leq x_{i}}&fg=000000$, by monotony

$latex \displaystyle \begin{array}{rl} F_{n}(x)-F(x) & \leq F_{n}(x_{i})-F(x_{i-1})\leq F_{n}(x_{i})-F(x_{i})+\frac{1}{k}\\ F_{n}(x)-F(x) & \geq F_{n}(x_{i-1})-F(x_{i})\leq F_{n}(x_{i-1})-F(x_{i-1})-\frac{1}{k} \end{array} &fg=000000$

Hence $latex {|F_{n}(x)-F(x)|\leq\sup_{i}|F_{n}(x_{i})-F(x_{i})|+1/k.}&fg=000000$ We finish, noticing that the first term goes to 0 (the $latex {\sup}&fg=000000$ is taken under a finite set).$latex \Box&fg=000000$



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