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How to calculate the following limit:

$$\lim_{C\rightarrow \infty} -\frac{1}{C} \log\left(1 - p \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!}\right)$$

Given that $0 \leq \gamma \leq 1$ and $0 \leq p \leq 1$. At least any tips about approaching the solution!

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You can use Poisson distributions. – Davide Giraudo Apr 7 '12 at 21:28
Could you give more details what do you mean? Actually, I want to find the exponent of a probability and poisson distribution is already envolved. – Osama Gamal Apr 7 '12 at 21:32
Consider a sequence $\{X_n\}$ of random variables which follow a Poisson distribution of parameter $\gamma$. Then $S_n:=\sum_{j=1}^nX_j$ follows a Poisson distribution of parameter $n\gamma$. If $\gamma=1$, central limit theorem gives that $\lim_{N\to \infty}\sum_{k=0}^n\frac{e^{-n\gamma}(n\gamma)k}{k!}=\frac 12$, and when $0\leq \gamma<1$ it converges to $1$. – Davide Giraudo Apr 7 '12 at 21:38
Similar assumption have been used in a previous somewhat related work here: (Appendix A). But, I can't apply the same approach. – Osama Gamal Apr 7 '12 at 21:58
up vote 4 down vote accepted

Note that $\mathrm e^{-\gamma C}\sum\limits_{k=0}^{C}\frac1{k!}(\gamma C)^k=\mathrm P(X_{\gamma C}\leqslant C)$, where $X_{\gamma C}$ denotes a Poisson random variable with parameter $\gamma C$. In particular, if $p\lt1$, the argument of the logarithm is between $1-p$ and $1$. Likewise, if $\gamma=1$, a central limit argument yields $\mathrm P(X_{C}\leqslant C)\to\frac12$. In both cases, the limit is zero.

From now on, assume that $p=1$ and that $\gamma\lt1$. One is interested in $$ 1-\mathrm e^{-\gamma C}\sum\limits_{k=0}^{C}\frac1{k!}(\gamma C)^k=\mathrm P(X_{\gamma C}\gt C). $$ Introduce some i.i.d. Poisson random variables $\xi$ and $\xi_k$ with parameter $1$ and, for every positive integer $n$, $\eta_n=\xi_1+\cdots+\xi_n$. Then, on the one hand $\eta_n$ is a Poisson random variable of parameter $n$ and on the other hand, for every $t\gt1$, the behaviour of $\mathrm P(\eta_n\gt tn)$ is described by a large deviations principle. More precisely, $$ \mathrm P(\eta_n\gt tn)=\mathrm e^{-nI(t)+o(n)},\quad\text{where}\ I(t)=\max\limits_{x\geqslant0}\left(xt-\log\mathrm E(\mathrm e^{x\xi})\right). $$ In the present case, $\log\mathrm E(\mathrm e^{x\xi})=\mathrm e^x-1$ hence $I(t)=t\log t-t+1$ for every $t\gt1$. Using this result for $n=\lfloor\gamma C\rfloor$ and $t=1/\gamma$, one gets $$ \lim\limits_{C\to+\infty}-\frac1C\log\mathrm P(X_{\gamma C}\gt C)=\gamma I(1/\gamma)=\gamma-1-\log\gamma. $$

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I'm familiar with this derivation. But, still I can't get why you assumed that p = 1. That's an important parameter in my argument. – Osama Gamal Apr 8 '12 at 4:49
I didn't assume that p=1, I proved (and @GEdgar did it as well) that for every p<1 the limit exists and is 0 (and then I dealt with the p=1 case). I understand you worked on this problem before posting it here (which is excellent) but please, try to READ the solutions and their PROOFS. – Did Apr 8 '12 at 6:51
I've posted a similar problem and I'd be glad if you could help:… – Osama Gamal Jun 25 '12 at 18:20

Perhaps not what you wanted to ask ... $$\begin{align} &0 \le \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!} \le 1, \\ &\log(1-p) \le \log\left(1 - p \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!}\right) \le \log 1 = 0 \\ &\lim_{C \rightarrow \infty} -\frac{1}{C} \log\left(1 - p \sum_{k=0}^{C}\frac{e^{-\gamma C} (\gamma C)^k}{k!}\right) = 0 \end{align}$$

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That's not true, it may be any value greater than $-\infty$ and less than $0$. – Osama Gamal Apr 7 '12 at 22:00
Only if $p=1$.. – GEdgar Apr 8 '12 at 3:04
Oh I got .. thanks – Osama Gamal Apr 8 '12 at 11:10

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