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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.

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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 $\ell=\frac23$.

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Yes, I was trying to get to that result the probabilistic way. – leonbloy Sep 13 '11 at 22:54

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