Probability

The Analysis of Data, volume 1

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 $r=1$.