Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
There are several equivalent definitions of convergence in distribution; which one(s) are you using? –  Nate Eldredge Oct 30 '12 at 13:57
    
I made some adjustments. Hope that helps. –  Christian Oct 30 '12 at 16:18

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.