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The question says $\{X_n\}^{\infty}_{n=1}$ and $X$ are real valued discrete random variables supported on the integers.

Given: $P(X_n = j) \Rightarrow P(X=j)$ for all integers $j$.

We are required to show that $X_n \Rightarrow X$, i.e. $X_n$ converges to $X$ in distribution.

My approach:

Fix $J \in \mathbb{Z}$. If I can show that $\sum\limits^{\infty}_{j=J} P(X_n=j) \rightarrow \sum\limits^{\infty}_{j=J} P(X=j)$ then we are done.

Let $F^k_n = \sum\limits^{k}_{j=J} P(X_n=j)$ and $F^k = \sum\limits^{k}_{j=J} P(X=j)$ for all $k \geq J$. Let $F_n = \sum\limits^{\infty}_{j=J} P(X_n=j)$ and $F = \sum\limits^{\infty}_{j=J} P(X=j)$.

$\therefore \lim\limits_{k \rightarrow \infty} F^k_n = F_n$ for all $n$, $\lim\limits_{n \rightarrow \infty} F^k_n = F^k$ for all $k$, and we have to prove that

$\lim\limits_{n \rightarrow \infty} \lim\limits_{k \rightarrow \infty}F^k_n = \lim\limits_{k \rightarrow \infty} \lim\limits_{n \rightarrow \infty}F^k_n = F$.

So basically we need to show that in this case we have sufficient conditions to interchange the order of limits. I tried doing that using ideas such as these, but haven't been able to make any headway yet. Any suggestion is greatly appreciated.

Canine360
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1 Answers1

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Since $\sum_{j=0}^{+\infty}P\left(X_n=j\right)=1=\sum_{j=0}^{+\infty}P\left(X=j\right)$, it suffices to show that $\sum_{j=0}^{J-1}P\left(X_n=j\right)\to \sum_{j=0}^{J-1}P\left(X=j\right)$ which follows from the assumptions.

That $\sum\limits^{\infty}_{j=J} P(X_n=j) \rightarrow \sum\limits^{\infty}_{j=J} P(X=j)$ implies the convergence in distribution is assumed by the opening poster.

Davide Giraudo
  • 172,925