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.

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!

share|improve this question
    
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: arxiv.org/pdf/1110.4703.pdf (Appendix A). But, I can't apply the same approach. –  Osama Gamal Apr 7 '12 at 21:58

2 Answers 2

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 $$ \lim\limits_{C\to+\infty}-\frac1C\log\mathrm P(X_{\gamma C}\gt C)=\gamma I(1/\gamma)=\gamma-1-\log\gamma. $$

share|improve this answer
    
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: math.stackexchange.com/questions/162941/… –  Osama Gamal Jun 25 '12 at 18:20

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

share|improve this answer
    
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

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.