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If N is a poisson random variable, why is the following true?

It is from "Probability and Stochastic Processes" by Yates, page 301, equation 8.2 to 8.3

$$ P\left(\left|\frac{N-E(N)}{\sigma_N}\right| \geq \frac{c}{\sigma_N}\right) = P(|Z| \geq \frac{c}{\sigma_N})$$

Z is the standard normal Gaussian random variable. The explanation in the text is : "Since E[N] is large", the CLT can be used. But I am familiar with the CLT being used with sums of random variables.

Thanks.

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The context here is telephone calls entering a system. The total number of calls is a poisson random variable, N. For example, if the hypothesis was the number of calls is 100, then I could express the observed random variable as $n_1 + n_2 + \cdots + n_{100}$. I think here, we are assuming the number of observed random variables is E[N]? And each random variable has expected value = variance = 1? Then, in general, is it true that each observation is poisson with expected value = variance = 1? –  jrand Apr 8 '11 at 4:00
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2 Answers 2

up vote 6 down vote accepted

Several points.

1) CLT only gives approximation to normal, not equality.

2) While the standard CLT can be easily applied to the case where the parameter $\alpha$ is an integer tending to $\infty$, you have a slight problem if $\alpha_n \to \infty$ with $\alpha_n$ real; consider Douglas Zare's comment: the sequence of parameters $\alpha _n /\left\lfloor {\alpha _n } \right\rfloor $ is not fixed (though tends to $1$).

3) This problem is essentially a special case of this recent one . Indeed, if $N$ has parameter $\alpha=t$, then it is equal in distribution to $X_t$, where $X = \{X_t: t \geq 0\}$ is a Poisson process with rate $1$. But $X$ is just a special case of a compound Poisson process, where the jump distribution is the $\delta_1$-distribution (this corresponds to the $Y_i$ being equal to $1$ in the linked post). So, instead of considering $\frac{{N - E(N)}}{{\sigma (N)}}$, you can consider $\frac{{X_t - E(X_t )}}{{\sigma (X_t )}}$ (which has been done in the linked post).

Remark. Note that in the linked post the $\frac{{X_t - E(X_t )}}{{\sigma (X_t )\sqrt {N_t } }} \to {\rm N}(0,1)$ appearing in question 1 should have been replaced with $\frac{{X_t - E(X_t )}}{{\sigma (X_t ) }} \to {\rm N}(0,1)$.

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N is a poisson($\alpha$) random variable. Then, it can be expressed as the sum of $\alpha$ poisson(1) random variables. If $\alpha$ is large, then the central limit theorem can be used. The reason for changing the poisson distribution is on page 253 of the text, the derivation involves the moment generating function.

$$ P\left(\left|\frac{N-E(N)}{\sigma_N}\right| \geq \frac{c}{\sigma_N}\right) $$ $$ = P\left(\left|\frac{\sum_{i=1}^{i=\alpha}n-\alpha}{\sqrt{\alpha}}\right| \geq \frac{c}{\sigma_N}\right) $$

Since $E(n) = {\sigma_n}^2 = 1$, set $Z_n = \frac{\sum_{i=1}^{i=\alpha}n-\alpha}{\sqrt{\alpha}}$ and $Z_n$ is standard normal, by CLT.

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What if $\alpha$ is not an integer? –  Mike Spivey Apr 8 '11 at 4:11
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Then use a sum of $\lfloor \alpha \rfloor$ IID Poisson distributions with parameter $\alpha/\lfloor \alpha \rfloor$. –  Douglas Zare Apr 8 '11 at 4:45
    
+1, thanks partly to the comment by Douglas Zare. –  Mike Spivey Apr 8 '11 at 5:12
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