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From binominal distribution if

  1. $n\to\infty$,
  2. $p\to 0$ and
  3. $n\cdot p=\lambda$ where λ constant

then binominal approaches poisson distribution. How can I define in typical form the above three conditions?

Calvin John
  • 572
Alexander Gotses
  • 1

1 Answers1

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Formally, you would say you have sequences $n_i\in\Bbb N$ and $p_i\in \Bbb R^+$ with

$$\lim_{i\to\infty} n_i=\infty,\qquad \lim_{i\to\infty} p_i=0, \qquad \lim_{i\to\infty} n_i p_i=\lambda$$

for some $\lambda \in\Bbb R$. To ensure that the third condition is always fulfilled you can exemplary choose

$$p_i:=\frac\lambda{n_i}.$$

In this case we always have $n_ip_i=\lambda$, and so this is also true in the limit.

M. Winter
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  • Sorry but can you explain better the third condition?How can i define the two sequences so the product is constant? Thanks – Alexander Gotses Jun 05 '17 at 18:58
  • @AlexanderGotses In this way? Or what do you want to know exactly? – M. Winter Jun 05 '17 at 19:11
  • I have seen that definition in some courses about probalities. The difficult part for me is the third condition. I can't understand how the product of the two sequences is constant. I i tried to define the sequences but i can not. Is it easy to define them ? If not can you point some source?? Thanks again. – Alexander Gotses Jun 05 '17 at 21:01
  • @AlexanderGotses I am not quite sure what you mean by defining them. I gave you definitions or formalizations but you do not seem to be satisfied. An example of such a pair of sequences would be $n_i\equiv 1,2,3,4,5,...$ and $p_i\equiv 1,\frac12,\frac13,\frac14,\frac15,...$. You see that $n_i\cdot p_i$ is always $1$, even though that $n$ is constantly increasing and $p$ is always decreasing. – M. Winter Jun 05 '17 at 21:13
  • Yes but as i remember and correct me if i'm wrong the product of ni *pi is not λ but is also a sequence . we define a third sequence λi wich is not a constant number . when n goes to infinity and p to zero then λi goes to λ. – Alexander Gotses Jun 05 '17 at 22:09
  • @AlexanderGotses Ok. Always keep in mind that the constant sequence $\lambda,\lambda,\lambda,...$ is also a sequence with the obvious limit $\lambda$. But take this example: $n_i\equiv 1,2,3,...$ and $p_i\equiv \frac12,\frac13,\frac14,...$, then $\lambda_i=n_i p_i\equiv \frac12,\frac23,\frac 34,...$ which will converge to $\lambda=1$. – M. Winter Jun 05 '17 at 22:38
  • The example is not successful as the product is a constant. Any source would be very helpful for me. The common books of probabilities i have assume the sequences. Thanks again . – Alexander Gotses Jun 05 '17 at 22:58
  • @AlexanderGotses I honestly have no idea what you are asking about exactly. All I can do is pointing you here. However, in my last example the product is no constant at all but depends on $i$. – M. Winter Jun 05 '17 at 23:01
  • Sorry i dint saw the last change you made. So if i want λ> 1 i just multiply the n sequence with λ. Thanks a lot . – Alexander Gotses Jun 05 '17 at 23:21