## Probability

### The Analysis of Data, volume 1

Slustky's Theorems

## 8.11. Slustky's Theorems

Proof. To prove the first statement, it is sufficient to show that for an arbitrary continuous function $h$ that is zero outside a closed and bounded set, $\E(h({\bb Y}^{(n)})) \to \E(h(X))$ (using the portmanteau theorem). Since a continuous function on a closed and bounded set in $\R^d$ (compact) is uniformly continuous, for all $\epsilon>0$ we can find $\delta>0$ such that $\|\bb x-\bb y\| < \delta$ implies $\|h(\bb x )-h(\bb y)\| < \epsilon$. Also, as a continuous function on a closed and bounded set in $\R^d$, $h$ is also bounded, say by $M$, and therefore \begin{align*} |\E(h({\bb Y}^{(n)})) - \E(h(X))| &\leq |\E(h({\bb Y}^{(n)})) - \E(h({\bb X}^{(n)}))| + |\E(h({\bb X}^{(n)})) - \E(h(X))|\\ &\leq |\E(h({\bb Y}^{(n)})) - \E(h({\bb X}^{(n)}))| I(\|{\bb X}^{(n)}-{\bb Y}^{(n)}\|\leq \delta) \\ &\quad + |\E(h({\bb Y}^{(n)})) - \E(h({\bb X}^{(n)}))| I(\|{\bb X}^{(n)}-{\bb Y}^{(n)}\|> \delta) \\ &\quad + |\E(h({\bb X}^{(n)})) - \E(h(X))|\\ &\leq \epsilon+2M \P(\|{\bb X}^{(n)}-{\bb Y}^{(n)}\|>\delta) + |\E(h({\bb X}^{(n)})) - \E(h(X))|. \end{align*} The second term in the equation above converges to 0 as $n\to\infty$ since $({\bb X}^{(n)}-{\bb Y}^{(n)})\toop \bb 0$. The third term also converges to zero since ${\bb X}^{(n)} \tood \bb x$. Since $\epsilon$ was arbitrarily chosen, it follows that $|\E(h({\bb Y}^{(n)})) - \E(h(X))|$, or $\E(h({\bb Y}^{(n)})) \to \E(h(X))$.

To prove the second statement, we note that if $\|{\bb Y}^{(n)}-\bb X\|>\epsilon$ occurs, then by the triangle inequality $\|{\bb Y}^{(n)}-{\bb X}^{(n)}\|+\|{\bb X}^{(n)}-\bb X\|>\epsilon$ occurs, implying that $\{\|{\bb Y}^{(n)}-\bb X\|>\epsilon\} \subset \{\|{\bb Y}^{(n)}-{\bb X}^{(n)}\|>\epsilon/2\} \cup \{\|{\bb X}^{(n)}-\bb X\|>\epsilon/2\}.$ This implies that \begin{align*} \P(\|{\bb Y}^{(n)}-\bb X\|>\epsilon) \leq \P(\|{\bb Y}^{(n)}-{\bb X}^{(n)}\|>\epsilon/2) + \P(\|{\bb X}^{(n)}-\bb X\|>\epsilon/2) \end{align*} which converges to 0 as $n\to\infty$ since ${\bb X}^{(n)} \toop \bb x$ and $({\bb X}^{(n)}-{\bb Y}^{(n)})\toop \bb 0$.

The third statement follows from arithmetic of deterministic limits, which apply since we have convergence with probability 1.

Proof. Since ${\bb Y}^{(n)}\tood \bb c$, we also have ${\bb Y}^{(n)}\toop \bb c$, implying that \begin{align*} \P(\| ({\bb X}^{(n)}, {\bb Y}^{(n)})- ({\bb X}^{(n)},\bb c) \| >\epsilon ) = \P(\|{\bb Y}^{(n)}-\bb c\|>\epsilon)\to 0. \end{align*} Let $h$ be an arbitrary continuous and bounded function. The statement $\E(h({\bb X}^{(n)},\bb c))\to \E(h(\bb X,\bb c))$ follows from the fact that ${\bb X}^{(n)}\tood \bb X$ and the portmanteau theorem. Combining this with Slutsky's theorem shows that $({\bb X}^{(n)},{\bb Y}^{(n)})\tood (\bb X,\bb c)$, which proves the first statement. To prove the second statement, we note that \begin{multline*} \P( \|({\bb X}^{(n)}, {\bb Y}^{(n)}) - (\bb X,\bb Y) \| >\epsilon) \leq \P(\|{\bb X}^{(n)}-\bb X \|>\epsilon/\sqrt{2})\\ + \P(\|{\bb Y}^{(n)}-\bb Y \|>\epsilon/\sqrt{2}), \end{multline*} and that the two terms in the right hand side above converge to 0 as $n\to\infty$. The third statement follows from arithmetic of deterministic limits, which apply since we have convergence with probability 1.
Corollary 8.11.2. If $\bb f$ is a continuous function, then ${\bb X}^{(n)}\tood \bb X \quad \text{ and }\quad {\bb Y}^{(n)}\tood \bb c\quad \text{imply}\quad \bb f ({\bb X}^{(n)},{\bb Y}^{(n)}) \tood \bb f({\bb X}^{(n)},\bb c).$
Example 8.11.1. If $X^{(n)}\tood X$ and $Y^{(n)}\toop c$ for some constant $c$, then
1. $X^{(n)} Y^{(n)}\tood Xc$
2. $X^{(n)}+Y^{(n)}\tood X+c$
3. $X^{(n)}/Y^{(n)}\tood X/c$,
where we assume in the last expression that $c\neq 0$.