Take the 2-minute tour ×
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.

Let $x_1, x_2 \cdots x_M$ be a sequence of iid random variables taking values over the integers, with $E(x_i)=0$. In particular, I'm interested in a shifted Poisson: $X=P-1$, where $P$ is Poisson with $\lambda=1$.

Let $y_1, y_2 \cdots y_M$ be the partial sums, $y_i=\sum_{k=1}^i x_k$

Finally, let $z_M$ be their uniform $mixture$: the result of choosing randomly, with equal probability $1/M$, one of the $y_i$ values. So that

$$p_{z_M}(t)=\frac{1}{M}\left[p_{y_1}(t)+p_{y_2}(t)+ \cdots +p_{y_M}(t)\right]$$

and $E[z_M]=0$. I'm interested in the asympotics of $p_{z_M}$, $M\to \infty$; in particular, I'd wish a good (better than first order) approximation of $P(z_M = 0)$.

I arrived to this at attacking (without much success) this problem, and got me interested about the general behaviour of $z_M$. An Edgeworth-like expansion does not seem feasible here, $z_M$ does not tend to a gaussian - its kurtosis does not tend to zero. It's easy to see that the characteristic function of $z$ is given by

$$Z(\omega) = \frac{X(\omega)}{M}\frac{1-{X(\omega)}^M}{1-X(\omega)}$$

but that did not lead me very far. Another observation: if we regard $y_M$ as a random walk (with Poisson steps), $P(z_M = 0)$ would equal the probality of finding the trajectory at $y=0$ at some random time.

share|improve this question
add comment

1 Answer

In the shifted Poisson case, the page you linked to indicates that $$ P(z_M=0)=\sqrt{\frac2{\pi M}}-\frac2{3M}+o\left(\frac1M\right). $$


Edit For every positive $k$, $y_k+k$ is Poisson with parameter $k$ hence $\mathrm P(y_k=0)=p_k$ with $$ p_k=\mathrm e^{-k}\frac{k^k}{k!}. $$ A refined form of Stirling's estimate yields $$ p_k=\frac1{\sqrt{2\pi k}}\mathrm e^{-\lambda_k},\qquad\qquad\mbox{with}\qquad\qquad\frac1{12k+1}<\lambda_k<\frac1{12k}, $$ hence $$ p_k=\int\limits_{k-1}^k\frac{\mathrm dt}{\sqrt{2\pi t}}-q_k-r_k, $$ with $$ q_k=\int\limits_{k-1}^k\left(\frac1{\sqrt{2\pi t}}-\frac1{\sqrt{2\pi k}}\right)\mathrm dt,\qquad\qquad r_k=\frac1{\sqrt{2\pi k}}(1-\mathrm e^{-\lambda_k}). $$ This yields $$ M\,\mathrm P(z_M=0)=\int\limits_{0}^M\frac{\mathrm dt}{\sqrt{2\pi t}}-\sum\limits_{k=1}^Mq_k-\sum\limits_{k=1}^Mr_k=\sqrt{\frac{2M}\pi}-\sum\limits_{k=1}^Mq_k-\sum\limits_{k=1}^Mr_k. $$ For every $t$ in $(k-1,k)$, $$ \frac1{\sqrt{t}}-\frac1{\sqrt{k}}=\frac{k-t}{\sqrt{t}\sqrt{k}(\sqrt{t}+\sqrt{k})}=O\left(\frac1{k^{3/2}}\right), $$ hence $$ q_k=O\left(\frac1{k^{3/2}}\right),\qquad r_k=O\left(\frac1{k^{3/2}}\right), $$ in particular the positive sequences $(q_k)$ and $(r_k)$ are summable and $$ \sum\limits_{k=M+1}^{+\infty}q_k=O\left(\frac1{\sqrt{M}}\right),\qquad \sum\limits_{k=M+1}^{+\infty}r_k=O\left(\frac1{\sqrt{M}}\right). $$ One gets finally $$ M\,\mathrm P(z_M=0)=\sqrt{\frac{2M}\pi}-\ell+O\left(\frac1{\sqrt{M}}\right),\qquad \ell=\sum\limits_{k=1}^{+\infty}(q_k+r_k), $$ and one knows that $\ell$ is positive and finite. However, obviously, other arguments are needed to show that $\ell=\frac23$.

share|improve this answer
    
Yes, I was trying to get to that result the probabilistic way. –  leonbloy Sep 13 '11 at 22:54
add comment

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.