# how to show convergence in probability imply convergence a.s. in this case?

Assume that $X_1,\cdots,X_n$ are independent r.v., not necessarily iid,

Let $S_n=X_1+\cdots+X_n$, suppose that $S_n$ converges in probability, how can we show that $S_n$ converges a.s.?

• I am not sure whether the following would help. If $A$ is the set of $\omega$ for which $S_n(\omega)$ converges, by Kolmogorov 0-1 law, $P(A)=0$ or $1$. Commented Dec 10, 2011 at 7:49
• See Theorem 5.3.4 in Kai Lai Chung, A course in probability theory, 3rd ed., Academic Press, 2001.
– Did
Commented Oct 31, 2012 at 20:53

## 1 Answer

We can use Ottaviani's inequality: if we put $M_k:=\max_{1\leqslant i\leqslant k}|S_i|$ and $S_{k,n}:=\sum_{i=k+1}^nX_i$, then for all $\varepsilon >0$ we have $$\min_{1\leqslant k\leqslant n}P(|S_{k,n}|\leqslant\varepsilon)P(|M_n|>2\varepsilon)\leqslant P(|S_n|>\varepsilon).$$

Put $A_m:=\sup_{k\in\mathbb N^*}|S_{m+k}-S_m|$ and $A:=\inf_{m\in\mathbb N^*}A_m$. We have $$\{\{S_n\}\mbox{ doesn't converge}\}=\{A\neq 0\}\subset\bigcup_{\varepsilon\in\mathbb Q^+}\bigcap_{m\in\mathbb N^*}\{A_m>\varepsilon\}$$ and $$\{A_m>\varepsilon\}=\bigcup_{r\in\mathbb N^*}\left\{\sup_{1\leqslant k\leqslant r}|S_{m+k}-S_m|>\varepsilon\right\}.$$ Ottaviani's inequality gives $$\min_{1\leqslant k\leqslant r}P(|S_{m+r}-S_{m+k}|\leqslant\varepsilon)P\left(\max_{1\leqslant k\leqslant r}|S_{m+k}-S_m|>2\varepsilon\right)\leqslant P(|S_{r+m}-S_m|>\varepsilon).$$ We fix $\delta>0$; we can choose $N=N(\varepsilon,\delta)$ such that for $0\leqslant k\leqslant r$ and $m\geqslant N$, $P(|S_{m+r}-S_{m+k}|>\varepsilon)\leqslant\delta$. We get $$P\left(\max_{1\leqslant k\leqslant r}|S_{m+k}-S_{m+r}|>\varepsilon\right)\leqslant\frac{\delta}{1-\delta},$$ therefore $P(A_m>\varepsilon)\leqslant\frac{\delta}{1-\delta}$. We have $$0\leqslant P\left(\bigcap_{p\in\mathbb N^*}(A_p>\varepsilon)\right)\leqslant P(A_m>\varepsilon)\leqslant\frac{\delta}{1-\delta},$$ so for all $\varepsilon\in\mathbb Q^+$, we have $P\left(\bigcap_{p\in\mathbb N^*}(A_p>\varepsilon)\right)=0$, hence $P(A\neq 0)=0$ and we have the almost sure convergence.

• great.....I tried without Ottavani.....Ottavani is necessary else we get a bound of N$\delta$ and cant control the probabilities Commented Feb 17, 2013 at 0:45