# Convergance of some series of random variables

Random variables $(X_{n})$ are independent and have the distribution $P(X_{n}=1)=p, P(X_n=-1)=1-p$, $\frac{1}{2}<p<1$. Prove that $$X_1+X_2+\dots+X_n \to \infty$$ almost sure.

Let $Y_n^{(a)}:=\begin{cases} X_n if \left|X_n \right| \le a\\ 0 otherwise \end{cases}$.

So this series don't satisfy the necessary contidion from Kolmogorov's three-series theorem (because $\sum_{n=1}^{\infty}\text{E}Y_{n}^{(1)}= \sum_{n=1}^{\infty}(2p-1)= \infty$, it must be $\sum_{n=1}^{\infty}X_{n}= \infty$.

Is this argument correct?

• Hint: Surely you know the behaviour when $n\to\infty$ of the sequence $$\frac{X_1+X_2+\cdots+X_n}n.$$ – Did Jun 13 '15 at 21:01
• It converges to 2p-1>0 – mrnobody Jun 13 '15 at 21:12
• So $X_{1}+\dots+X_{n} \ge (2p-1)n$ almost sure – mrnobody Jun 13 '15 at 21:16
• And having $2p-1 > 0$ we can conclude that $X_1+\dots+X_n \to \infty$ – mrnobody Jun 13 '15 at 21:20
• @sebuss007 Why not write it as an answer to your question? – saz Jun 14 '15 at 5:29

My solution:

Because $(X_n)$ are iid and $\text{E}\left|X_n \right|=1$ from SLLN we have:

$$\overline{X_n} \to \text{E}X_n=2p-1$$

Let $0<q< 2p-1$. So there almost sure is:

$$X_1+X_2+\dots+X_n > qn$$

And having $q>0$ we can conclude that $X_1+X_2+\dots+X_n \to \infty$ a.s.

• No, $S_n/n\to2p-1$ does not imply that $S_n\geqslant(2p-1)n$ for every $n$ large enough. Forget the randomness, how would you proceed to show that a deterministic sequence $(s_n)$ such that $s_n/n\to\ell$ with $\ell>0$, converges to $+\infty$? – Did Jun 14 '15 at 11:21
• But doesn't it also imply that $S_n \ge (2p-1)n$ almost sure? – mrnobody Jun 14 '15 at 11:34
• Did you read my previous comment? Anyway, imagine $s_n=(2p-1)n-42$ for every $n$. Then what? – Did Jun 14 '15 at 11:37
• Oh, I see, you're right, but if we choose some number between 0 and 2p-1, everything will be ok, won't it? – mrnobody Jun 14 '15 at 11:50
• This is true.  – Did Jun 14 '15 at 11:58