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Let $X_n$ with $n\in[1,+\infty ]$ r.v. taking value in $\mathbb Z$. I want to show that if $\mathbb P\{X_n=m\}\longrightarrow \mathbb P\{X_\infty =m\}$ then $X_n\longrightarrow X_\infty $ in distribution. There is things I don't understand in the proof.

Proof

Suppose $\mathbb P\{X_n=m\}\longrightarrow \mathbb P\{X_\infty =m\}$.Let $x\in \mathbb R$ and $\varepsilon>0$. Let $M\in\mathbb Z$ s.t. $M>x$ s.t. $$\mathbb P\{X_\infty \notin[-M,M]\}<\varepsilon.$$ Then, choose $N\in\mathbb N$ s.t. $n\geq N$ implies $$\sum_{i=-M}^M|\mathbb P\{X_n=i\}-\mathbb P\{X_\infty =i\}|<\varepsilon.$$ Then, when $n\geq N$,

$$\mathbb P\{X_n\notin [-M,M]\}=1-\sum_{i=-M}^M\mathbb P\{X_n=i\}\underset{\color{red}{(1)}}{\leq} 1-\sum_{i=-M}^M \mathbb P\{X_\infty =i\}+\varepsilon \underset{\color{red}{(2)}}{\leq} 2\varepsilon.$$

Question 1 : Where does $(1)$ and $(2)$ come from. I spent lot of time on it but I still don't understand.

Then, if $n\geq N$ $$|F_{X_\infty }(x)-F_{X_n}(x)|=\left|\sum_{i=-\infty }^{\lfloor x\rfloor}\mathbb P\{X_\infty =i\}-\mathbb P\{X_n=i\}\right|$$

$$\leq \sum_{i=-\infty }^{-M-1}\mathbb P\{X_{\infty }=i\}+\sum_{i=-\infty }^{-M-1}\mathbb P\{X_n=i\}+\sum_{i=-M}^M|\mathbb P\{X_{\infty }=i\}-\mathbb P\{X_n=i\}|\leq 4\varepsilon.$$

Question 2 : I agree that $\sum_{i=-M}^M|\mathbb P\{X_{\infty }=i\}-\mathbb P\{X_n=i\}|<\varepsilon$, but I don't understand at all why $$\sum_{i=-\infty }^{-M-1}\mathbb P\{X_{\infty }=i\}+\sum_{i=-\infty }^{-M-1}\mathbb P\{X_n=i\}\leq 3\varepsilon.$$ I really don't see where it come from.

1 Answers 1

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For your first question,

$A-B\leq |A-B|$ and $B-A\leq |A-B|$. So (1) can be shown by throwing $\sum_{i=-M}^MP(X_\infty=i)$ to the left-hand-side. For (2), look at your assumptions above, and realize $1-\sum_{i=-M}^MP(X_\infty=i)=P(X_\infty\notin [-M,M])$.

For your second question, each sum is bounded above by $P(X_\infty\notin [-M,M])$ and $P(X_n\notin[-M,M])$ respectively.