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Why do many “Introduction to Probability”-books say:…when n is large and p is small … is approximately Poisson np,Why should n be large?

Let Y_i be a bernoulli P[Y_i = 1] = p,Let M = \sum_{i=1}^{n} = Y_i, M is distributed Binomial:P[M = x] = n!/(x!(n-x)!) p^x (1-p)^{n-x}.Now, let np = z then:P[M =x] = (z/n)^x/x! [1-(z/n)]^{n-x} n!/(n-x)!,P[M =x] = z^x/x! [1-(z/n)]^{n-x} n(n-1)...(n-x+1)/n^x,P[M =x] = z^x/x! [1-(z/n)]^n/[1-(z/n)]^x n(n-1)...(n-x+1)/n^xRemark that when n tends to infinity:1) lim n(n-1)...(n-x+1)/n^x = 1 2) lim [1-(z/n)]^n = lim exp[n ln(1-(z/n))] = exp(-z)3) lim [1-(z/n)]^x = 1You've got the desired result.Hope this helps

Fasturtle, you are assuming that when n increases then p decrease. But that doesn't have to be the case. See for example the thread Probability of 8 Occurences?. There the Poisson Approximation is closer for n=1.000.000 then 200.000.000.The Poisson Approximation is pretty good for small fixed p and gets worse when n increase. So, what is the condition really, for a good approximation?

the condition is n is big, p is small(say p less than 0.01) and n*p is not to big not too small QuoteOriginally posted by: freyziFasturtle, you are assuming that when n increases then p decrease. But that doesn't have to be the case. See for example the thread Probability of 8 Occurences?. There the Poisson Approximation is closer for n=1.000.000 then 200.000.000.The Poisson Approximation is pretty good for small fixed p and gets worse when n increase. So, what is the condition really, for a good approximation?

Basically, to derive the Poisson as the limit of the binomial, you take the limit of the binomial distribution as n goes to infinity and p goes to zero in such a way that np is always constant.

For a Poisson distribution, the chance of getting k+1 successes is m/(k+1) times the chance of getting k successes, where m is the mean of the distribution.For a binomial distribution, the chance of getting k+1 successes is (n - k)*p/[(k+1)*(1-p)] times the chance of getting k successes. This can be rewritten as [m/(k+1)]/(1-p) - k*p/[(k+1)*(1-p)], and further as m/(k+1) + p*(m - k)/[(k+1)*(1-p)]. As long as I'm interested in reasonably high probabilities, (m - k) will be on the order of the square root of m. As long as p/(1-p) is small compared to the square root of m, the second term can be neglected and the binomial will be almost the same as a Poisson.p/(1-p) << m^0.5 implies p^0.5/(1-p) << n^0.5 or p/(1-p)^2 << n. p doesn't have much to do with this, unless it is near 1. If n is large, and p is not near 1, it will be true.

Yes, I agree with you that a fixed Poisson () distribution is the limit of a sequence of binomial distributions with parameters n and /n as n tends to infinity.But, what I am fishing for is:A fixed binomial distribution with parameters n and p can be approximated by a Poisson distribution with parameter np.The preciseness of this approximation decreases as n increases (assuming p fixed).Aaron is working in the right direction by comparing the distance between the two distributions. But I am sticking to “n large is bad”and p small (well not twice as good but) square as good I admit this might seem as a complicated result, but I am looking for how widely it is known. It can be proven nicely by using coupling.Anyone know the result?

Let be the sum of n independent Bernoulli random variables with thenwith . So, preciseness is decreasing in n.For proof, see LeCam's Inequality and Poisson Approximation, by Mike Steele.