Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could anyone help to solve this problem in probability? Thank you very much!

Let $\{X_{k}\}_{k\in N}$ be iid $0$-$1$ random variables. Decide the asymptotic behavior of the random walk $S_{n}:=\sum\limits_{k=1}^{n}\frac{1}{k}X_{k}$.

You may begin with showing that the tail of the random walk decays to zero. Namely $\forall \epsilon>0$, $$ \Pr [ |T_{k}|>\epsilon \text{ infinite often}]=0 ,$$ where $T_{k}:=\sum\limits_{i=k}^{\infty}\frac{1}{i}X_{i}$.

share|cite|improve this question
Is this homework? What have you tried? – Srivatsan Sep 9 '11 at 19:06
Are you sure the values are supposed to be $0$ and $1$, not $-1$ and $1$? As it is, $T_k = \infty$ almost surely (except in the trivial case where all $X_k = 0$ almost surely). – Robert Israel Sep 9 '11 at 19:12
@Robert I don't think $\pm 1$ alone is enough; shouldn't they be zero mean random variables as well? – Srivatsan Sep 9 '11 at 19:19
Perhaps a bit nitpicky, but I do not understand why $T_k$ is indexed with $k$, when the $k$ is already used for the $X_k$. Writing $T_n = \sum_{k=n}^{\infty} \frac{1}{k} X_k$ would fit better with the notation $S_n$ and $X_k$. – TMM Sep 9 '11 at 19:21
@Srivatsan: yes, of course, but the point is that with $0$ and $1$ there is no nontrivial case that works. Actually, I suspect that what was meant was $N(0,1)$, i.e. normal with mean $0$ and variance $1$. – Robert Israel Sep 10 '11 at 1:51
up vote 3 down vote accepted

Assume that $\mathrm P(X_n=1)=x$ and $\mathrm P(X_n=0)=1-x$, hence $\mathrm E(X_n)=x$ for every $n$, and let $H_n=\sum\limits_{k=1}^n\frac1k$ denote the $n$th harmonic number, hence $H_n=\log(n)+\gamma+o(1)$.

Then $S_n-xH_n$ converges almost surely and in $L^2$ to an almost surely finite centered random variable $Y$ with variance $\mathrm E(Y^2)=x(1-x)\frac{\pi^2}6$.

In particular, $\frac1{\log n}S_n\to x$.

To see this, consider $Y_n=\sum\limits_{k=1}^n\frac1k(X_k-x)$. The random variables $X_k-x$ are centered, square integrable with variance $x(1-x)$, and independent. Hence, for every $n$, $Y_n$ is centered with variance $\sum\limits_{k=1}^n\frac1{k^2}\mathrm E((X_k-x)^2)=\sum\limits_{k=1}^n\frac1{k^2}x(1-x)$. The series $\sum\limits_k\frac1{k^2}x(1-x)$ converges hence $(Y_n)$ converges in $L^2$. Since $(Y_n)$ is the sequence of the partial sums of some independent random variables, a result due to Paul Lévy ensures that $(Y_n)$ converges almost surely as well.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.