Limit Theorems

## 8.3. Dominated Convergence Theorem for Random Vectors

We have the following extensions of the Dominated Convergence Theorem (see Chapter F).

Proposition 8.3.1 (Dominated Convergence Theorem for Random Variables). Consider a sequence of RVs $X^{(n)}$ for which $|X^{(n)}|\leq Y$ for some RV $Y$ with $\E|Y| < \infty$. Then $X^{(n)}\tooas X \qquad \text{implies} \qquad \E(X^{(n)}) \to \E(X).$
Proof. We construct a discrete-time random process $\mathcal{X}=(X^{(n)}, n\in\mathbb{N})$ and consider the Dominated Convergence Theorem with $\mu$ being the distribution of the random process $\mathcal{X}$, $f_n=X^{(n)}$, and $f=X$ (see Proposition F.3.9). In this case $\int f_n\,d\mu$ becomes $\E(X^{(n)})$ and $\int f\,d\mu$ becomes $\E(X)$.
Proposition 8.3.1 (Dominated Convergence Theorem for Random Vectors). If ${\bb X}^{(n)} \tooas \bb X$ and $\|\bb X\|^r\leq Y$ for some RV $Y$ with $\E Y < \infty$ and some $r>0$, then $\E(\|{\bb X}^{(n)}-\bb X\|^r)\to 0.$
Proof. ${\bb X}^{(n)} \tooas \bb X$ and $\|\bb X\|^r\leq Y$ implies $\|\bb X^{(n)}\|\leq Y$ with probability 1, which implies $\|{\bb X}^{(n)}-\bb X\|^r \leq (\|{\bb X}^{(n)}\| + \|\bb X\|)^r \leq (Y^{1/r}+Y^{1/r})^r=2^r Y$ with probability 1. The result then follows from applying the dominated convergence theorem for random variables to the sequence of RVs $\|{\bb X}^{(n)}-\bb X\|^r, n\in\mathbb{N}$ (that sequence converges with probability 1 to 0 since ${\bb X}^{(n)}\tooas X$).

The name of the proposition above is motivated by the fact that convergence $\E(\|{\bb X}^{(n)}-\bb X\|^r)\to 0$ is sometimes called convergence in the $r$-mean and denoted ${\bb X}^{(n)}\to^r \bb X$. The dominated convergence theorem for random vectors implies convergence in $r$-mean of the vector ${\bb X}^{(n)}$ to the vector $\bb X$, which is similar to the dominated convergence theorem for random variables for $r=1$.