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The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]

Let n -> infinity while np = L, a constant, so that p = L/n

then

P(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)

= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)

= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)

= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)


Now lim n -> infinity of (1 - L/n)^n = e^(-L)

and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1

lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)


So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.

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Q: Why is it said that poisson distribution is a limiting case of binomial distribution?
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Derive normal distribution as a limiting case of binomial distribution?

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The mean and variance are equal in the Poisson distribution. The mean and std deviation would be equal only for the case of mean = 1. See related link.


What is a method for multiplying two binomials?

Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.


Is (a-3) a the sum of two terms?

Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.


What does lambda mean in maths?

In the ancient Greek system, the letter Lambda had a value of 30. Nowadays, the lower case lambda stands for the wave length. In statistics, it is the parameter of the Poisson distribution and represents the reciprocal of the frequency.

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Derive normal distribution as a limiting case of binomial distribution?

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The mean and variance are equal in the Poisson distribution. The mean and std deviation would be equal only for the case of mean = 1. See related link.


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This depends on what information you have. If you know the success probability and the total number of observations, you can use the given formulas. Most of the time, this is the case. If you have data or experience which allow you to estimate the parameters, it may sometimes happen that you work like this. This mostly happens when n is very large and p very small which results in an approximation with the Poisson distribution.


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Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.


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Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.


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