Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X_1, X_2, X_3,\dots$ be an i.i.d. sequence of random variables with finite mean. Write $S_n=X_1+X_2+\dots+X_n$.

Let $N$ be a non-negative integer-valued random variable with finite mean. $N$ may not be independent of the sequence $(X_i)$.

Is it necessarily the case that $S_N$ has finite mean?

Of course, it's true if $N$ is independent of the sequence $(X_i)$. Then $E(S_N)=E(N)E(X_1)$. It's still true if $N$ is a stopping time for the sequence $(X_i)$.

It's also true if the $X_i$ have finite variance. Then for any $c>E(X_i)$, the quantity $R_c=\sup(S_n-cn)$ has finite mean, and $E(S_N)\leq cE(N)+E(R_c)$.

share|improve this question

3 Answers 3

up vote 11 down vote accepted

Edit: I made it a bit clearer and simpler.

No. You start with noticing that there is no "linear" estimate for the mean of $S_N$ in terms of the mean of the sample $X$ under the assumption that the mean of $N$ is small. To this end just take $X$ be $0$ with probability $1-p$ and $A$ with probability $p$ so that $Ap$ is small. Now, once we have a sequence $X_i$ of independent copies of $X$, define $N$ to be $m$ if at least one of $X_1,\dots,X_m$ is not $0$ and $0$ otherwise. Then $ES_N= Apm$ and $EN\le m^2p$ and $EX=Ap$.

Now choose $A_j,p_j,m_j$ so that $\sum A_jp_j<+\infty$, $\sum A_jp_jm_j=+\infty$, $\sum m_j^2p_j<+\infty$. For instance, take $m_j=2^j$, $p_j=2^{-3j}$, $A_j=2^{2j}$.

Define the random samples $X^{(j)}$ and random interval lengths $N^{(j)}$ as above using $A_j,p_j,m_j$ in place of $A,p,m$. Put $X=\sum_j X^{(j)}$, $N=\sum_j N^{(j)}$. Then $EX$ and $EN$ are finite. Let $X_i$ be independent copies of $X$. We have $E\sum_1^N X_i\ge E\sum_j \sum_1^{N^{(j)}}X_i^{(j)}=\sum_j A_jp_jm_j=+\infty$.

share|improve this answer
    
I don't understand. Are your $X_i$ iid? If so, then what does the index $j$ refer to in the second-to-last sentence? –  Louigi Addario-Berry Feb 22 '11 at 19:50
    
I mean, you choose $X^{(j)}$ and $N^{(j)}$ the way I described (the $X^{(j)}$ are different samples, not the different copies of each sample) and then put $X=\sum_j X^{(j)}$ (that is just one sample which you use for making copies) and $N=\sum_j N^{(j)}$. –  fedja Feb 22 '11 at 20:04
    
Thanks fedja - nice argument. In fact, for any $\epsilon$ we can arrange it that $\sum A_j^{2-\epsilon} p_j<\infty$. I think this would give an example with a distribution $X$ satisfying $E X^{2-\epsilon}<\infty$. So it sounds like the finiteness of the variance (or something close to it) may actually be the relevant condition here. –  James Martin Feb 25 '11 at 10:13

Here's a counterexample.

Let $X$ be equal to $2^k k^{-2}$ with probability $2^{-k}$. The probability that among $n$ i.i.d. copies of $X$ we get at least one with value $2 ^ {2 \log n} (2 \log n)^{-2}= n^2 (2 \log n)^{-2}$ is about $n^{-1}$. Call this event $A_n$.

Thus, if we choose $N$ to be $2^{2k/3}$ with probability $2^{-k}$ such that the event $N=2^k$ is a subset of $A_{2^{2k/3}}$, we get that $S_N$ have infinite expectation. This should be quite easy to do since the events $A_n$ tend to be independent for far away $n$'s.

share|improve this answer

Assume $X_1$ has infinite variance and let $N$ denote the first time $n$ such that $S_n-cn=R_c$. By a result I do not manage to find a reference for but that you surely know, $R_c$ is not integrable hence $S_N$ is not either. To get an answer to your question, it would remain to prove that $N$ is integrable.

The probability that $N$ is not one of the $n$ first times of record should decrease geometrically hence if the time of a first record is integrable, conditionally on the fact that there is a positive record, we are done. I seem to remember that a random walk with a negative drift conditioned on hitting the positive halfline is a random walk based on different increments with a positive drift. One should be able to truncate them to get bounded increments still with a positive drift. Then the result becomes trivial. (Not sure this whole reasoning is watertight, though. If it is not, shoot.)

share|improve this answer
1  
If $X_1$ has infinite variance we can arrange it so that $P(X_1 > n) \asymp 1/(n \log n (\log\log n)^2)$, say, and then the probability that one of $X_n,...,X_{2n}$ has value at least $2cn$, say, is of order $1/(\log n (\log\log n)^2)$. This doesn't quite shoot down the approach but it does show that $N$ need not decrease geometrically. –  Louigi Addario-Berry Feb 22 '11 at 19:16
    
@Louigi Surely this is me but how does this apply? –  Did Feb 22 '11 at 20:00
    
@Didier, do you mean how does my choice of distribution for $X_1$ apply to James' question? I'm not sure I understand your question. At any rate, I think you are probably right that N should be integrable in your construction, I was just trying to say that it can have a much heavier tail than geometric. –  Louigi Addario-Berry Feb 22 '11 at 20:53
    
@Louigi Call $q_n$ the probability that at least one $X_i$ for $i$ between $n$ and $2n$ is at least $2cn$. Call $p_n$ the probability that $N$ is not one of the first $n$ record times (of the random walk with steps distributed like $X_1-c$). I fail to see why a lower bound on $q_n$ implies a lower bound on $p_n$. (As an aside, note that $p_n$ thus defined is not the probability that $N\ge n+1$.) –  Did Feb 22 '11 at 21:20
    
@Didier, you're right, they're not related, I just wasn't reading what you wrote properly. Sorry! –  Louigi Addario-Berry Feb 22 '11 at 22:36

Your Answer

 
discard

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.