Question

Let $\Psi=\Psi_1 \times \Psi_2 \times \ldots$ be a countable product of Polish spaces, as in Theorem 19, and let $\left(Q_n\right)$ be a sequence of probability measures on $\Psi$. Show that $\left(Q_n\right)$ is uniformly tight if and only if $\left(Q_n^j\right)$ is uniformly tight for each $j$, where $Q_n^j$ is the $j^{\text {th }}$ coordinate marginal of $Q_n$.

   Let $\Psi=\Psi_1 \times \Psi_2 \times \ldots$ be a countable product of Polish spaces, as in Theorem 19, and let $\left(Q_n\right)$ be a sequence of probability measures on $\Psi$. Show that $\left(Q_n\right)$ is uniformly tight if and only if $\left(Q_n^j\right)$ is uniformly tight for each $j$, where $Q_n^j$ is the $j^{\text {th }}$ coordinate marginal of $Q_n$.

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A Modern Approach to Probability Theory (Probability and Its Applications)

A Modern Approach to Probability Theory (Probability and Its Applications)

Bert E. Fristedt,… 1st Edition

Chapter 18, Problem 16

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Step 1

A sequence of probability measures \((Q_n)\) on a Polish space \(\Psi\) is uniformly tight if for every \(\epsilon > 0\), there exists a compact set \(K \subset \Psi\) such that \(Q_n(K) > 1 - \epsilon\) for all \(n\). Show more…

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Let $\Psi=\Psi_1 \times \Psi_2 \times \ldots$ be a countable product of Polish spaces, as in Theorem 19, and let $\left(Q_n\right)$ be a sequence of probability measures on $\Psi$. Show that $\left(Q_n\right)$ is uniformly tight if and only if $\left(Q_n^j\right)$ is uniformly tight for each $j$, where $Q_n^j$ is the $j^{\text {th }}$ coordinate marginal of $Q_n$.

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