5
$\begingroup$

I need help proving the Continuous Mapping Theorem (CMT) for random vectors. I'm currently reading Econometric Analysis for Cross Section and Panel Data by Jeffrey M. Wooldridge (Chapter 3, pp. 40 - 41, 2nd edition). Unfortunately, he leaves it to the reader to prove most asymptotic results. Additionally, almost every other econometrics textbook I read simply states the result.

Definition 1: A sequence of random variables $x_n$ converges in distribution to a continuous random variable $x$ if and only if $\forall s \in \mathbb{R} \ \forall \epsilon >0 \ \exists N \ s.t. \ \forall n>N \; |Prob(x_n \leq s) - Prob(x \leq s)|<\epsilon$. We write $x_n \to^d x.$ [Note: A continuous random variable is one for which the cumulative distribution function is continuous.]

Definition 2: A sequence of K $\times$ 1 random vectors $\mathbf{x}_n$ converges in distribution to the continuous random $K \times 1$ vector $\mathbf{x}$ if and only if $\forall \mathbf{c} \in \mathbb{R}^{K}$ such that $\mathbf{c}^T\mathbf{c} = 1$, $\mathbf{c}^T\mathbf{x}_n \to^d \mathbf{c}^T\mathbf{x}$, and we write $\mathbf{x}_n \to^d \mathbf{x}.$

Theorem 1: Let $\mathbf{x}_n$ be a sequence of $K \times 1$ random vectors such that $\mathbf{x}_n \to^d \mathbf{x}$. If $\mathbf{g}:\mathbb{R}^k\to\mathbb{R}^{\ell}$ is a continuous function, then $\mathbf{g}(\mathbf{x}_n)$ $\to^d$ $\mathbf{g}(\mathbf{x}).$

Definition 3: A sequence of random variables $x_n$ is bounded in probability if and only if $\forall \epsilon>0 \ \exists b_{\epsilon}>0 \ \exists N \ s.t. \forall n>N \ Prob(|x_n|>b_{\epsilon})$. A vector $\mathbf{x}_n$ is bounded in probability if and only if the random variables which constitute the vector of random variables are themselves bounded in probability.

Theorem 2: If $\mathbf{x}_n \to^d \mathbf{x}$, where $\mathbf{x}$ is a $K \times 1$ vector, then $\mathbf{x}_n = O_p(1)$.

I need rigorous proofs for Theorems 1 and 2. This problem has been frustrating me for a couple days now, so any help would go a long way.

Thanks.

CS

$\endgroup$
2
  • $\begingroup$ There are several equivalent definitions of convergence in distribution; which one(s) are you using? $\endgroup$ Oct 30, 2012 at 13:57
  • $\begingroup$ I made some adjustments. Hope that helps. $\endgroup$
    – Christian
    Oct 30, 2012 at 16:18

1 Answer 1

3
$\begingroup$

For Theorem 1:

Let $x_n$ be defined on the probability space $(\Omega, \mu)$. Your Definition 2 for convergence in probability of a sequence of random vectors says that for any half space $H$ of $\mathbb{R}^k$, i.e. $H = \phi^{-1}(r)$ for some linear functional $\phi: \mathbb{R}^k \rightarrow \mathbb{R}$ and $r \in \mathbb{R}$, $\mu(x_n^{-1}(H)) \rightarrow \mu(x^{-1}(H))$. ($\phi$ is inner product with $c$ in your definition.)

Now if $g: \mathbb{R}^k \rightarrow \mathbb{R}^l$ is linear, then Theorem 1 is immediate: For any half space $H \subset \mathbb{R}^l$, $g^{-1}(H)$ is again a half space of $\mathbb{R}^k$. So $\mu(x_n^{-1}(g^{-1}(H))) \rightarrow \mu(x^{-1}(g^{-1}(H)))$.

The case $g$ is just measurable takes a little doing. Definition 2 implies the following: for any closed convex $C \subset \mathbb{R}^k$, $\mu(x_n^{-1}(C)) \rightarrow \mu(x^{-1}(C))$. This can be shown by writing $C$ as the countable intersection of polygons and use continuity-from-above of the pushforward measures. Now take any half space $H \subset \mathbb{R}^l$. Consider the measurable set $g^{-1}(H)$. The pushforward measure $\mu$ induced by $x$ is regular. So it can be approximated from below by some compact $K \subset g^{-1}(H)$. In turn, $K$ can be covered by finite rectangles $C_1,\cdots, C_m$. Since $\mu(x_n^{-1}(C_i)) \rightarrow \mu(x^{-1}(C_m))$ for $i = 1,\cdots,m$, Theorem 1 holds.

For Theorem 2:

Let $B_b$ denote the closed cube centered at the origin of radius $b$ in $\mathbb{R}^k$.

Your Definition 3 says that, for all $\epsilon > 0$, there exists $b> 0$ and $N \in \mathbb{N}$ such that $\mu(x_n^{-1}(B_b)) > 1 - \epsilon$ for all $n \geq N$.

$B_b$ is convex and closed. So $\mu(x_n^{-1}(B_b)) \rightarrow \mu(x^{-1}(B_b))$ for any such cube. By the regularity of the pushforward measure again, $\mu(x^{-1}(B_b)) \rightarrow 1$ as $b \rightarrow \infty$. So Theorem 2 holds.

I am an econ grad student myself. Hope this helps.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .