Simple random walk in the limit

Question

Consider simple random walk: $X_n=\pm1$ with equal probabilities. $S_n =\sum_{i=1}^nX_i$. For finite $n$ we can write

$$S_n=\sum_{i=1}^nX_i=\sum_{i=1}^nX_i^+ -\sum_{i=1}^nX_i^-$$ So that $$E[S_n]=\sum_{i=1}^nE[X_i^+] -\sum_{i=1}^nE[X_i^-]$$

However, $\lim_{n\rightarrow\infty}\sum_{i=1}^nE[X_i^+]=\infty$ and $\lim_{n\rightarrow\infty}\sum_{i=1}^nE[X_i^-]=\infty$.

So, what happens to $E[S_n]$ in the limit?

score 1 · Accepted Answer · answered Oct 19 '14 at 22:44

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Simply we have $$E(S_n) = E(\sum_{i=1}^n X_i) =\sum_{i=1}^n E( X_i) = \sum_{i=1}^n 0 = 0$$ so $\lim_{n\to\infty}E(S_n) = \lim_{n\to\infty} 0 = 0$

answered Oct 19 '14 at 22:44

Petite Etincelle

You are using linearity of expectations. Actually, my question is if it holds in the limit in this case? – Oct 19 '14 at 22:52
@d.k.o. If all $ES_n$ are $0$, then its limit is still zero. what you asks for is $\lim E(S_n)$, not $E(\lim S_n)$ – Petite Etincelle Oct 19 '14 at 22:54
OK. So, are $\lim E(S_n)$ and $E(\lim S_n)$ equal in this case? I mean, Fubini does not help here because $\sum_{n=1}^\infty E[|X_n|]=\infty$ – Oct 19 '14 at 22:55
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@d.k.o. $\lim S_n$ does not exist, we can only say $\limsup_n S_n = +\infty$ and $\liminf_n S_n = -\infty$ – Petite Etincelle Oct 19 '14 at 22:59
Thank's. It seems trivial, but I was looking for a formal justification... – Oct 19 '14 at 23:00
@d.k.o. for a formal justification, maybe you can look at this – Petite Etincelle Oct 19 '14 at 23:13

1 Answers1