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.

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.


share|improve this question
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

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 $\frac{{X_t - E(X_t )}}{{\sigma (X_t ) }} \to {\rm N}(0,1)$.

share|improve this answer

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.

share|improve this answer
What if $\alpha$ is not an integer? –  Mike Spivey Apr 8 '11 at 4:11
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

Your Answer


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.