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ST2334: SOME NOTES ON THE BINOMIAL AND POISSON DISTRIBUTIONS Binomial Distribution We first note the binomial coefficient: the number of ways, disregarding order, that x objects can be chosen from amongst n objects: n n! = . x (n − x)!x! Let n ≥ 1 be fixed and finite integer and let X = {0, 1, . . . , n}. Then a random variable X has a binomial distribution, if its PMF is: n x p (1 − p)n−x x ∈ X, p ∈ (0, 1). f (x) = x Such a PMF is suitable for a random variable that counts the ‘successes’ in a collection of n independent and identical Bernoulli trials (an experiment where there are only two possible outcomes), where the probability of success is p ∈ (0, 1). An example of which, can be found in the notes; flipping a coin n times which lands heads with probability p and counting the number of heads which lands. In this example, the Bernoulli trial is that of flipping the coin; there are only two possible outcomes, head or tail. The success is a ‘heads’ landing and one would expect that these consecutive Bernoulli trials are independent. We remark that f (x) is indeed a valid PMF. First f (x) ≥ 0 ∀x ∈ X. This is because 1 > p > 0 and so for any x ∈ X, px (1 − p)n−x > 0. The binomial coefficient is always positive in our scenario. Second, n n X X n x f (x) = p (1 − p)n−x = 1. x x=0 x=0 Here we use the fact that for any fixed a, b ∈ (−∞, ∞) and poitive integer n, n X n x n−x a b . (a + b)n = x x=0 In our setting a = p and b = 1 − p and for any n ≥ 1, (1)n = 1. Thus, our PMF is such that 1 ≥ f (x) ≥ 0 for each x ∈ X. The distribution function is for x ∈ X x X n y P(X ≤ x) = F (x) = p (1 − p)n−y . y y=0 This latter summation does not always have an analytic expression. One can use the distribution function to answer various questions of interest. For example; if one flips a coin 10 times and the probability of heads is p, what is the probability of: 1 2 ST2334 • 1 head • at least 9 heads For the first question we have P(X = 1) = f (1) = F (1) − F (0) = 10 p(1 − p)9 . 1 For the second question, this is 10 9 10 10 P(X ≥ 9) = f (9) + f (10) = F (10) − F (8) = p (1 − p) + p . 9 10 Poisson Distribution Let X be the set of non-negative integers: X = {0, 1, . . . }. Then a random variable X ∈ X is said to have a Poisson distribution of parameter ∞ > λ > 0 if the PMF is λx e−λ f (x) = x ∈ X. x! In general, the Poisson distribution is often used for random variables associated to counts (for example, counts which can grow indefinitely). This is a PMF as f (x) ≥ 0 ∀x ∈ X: λx /x! > 0 and e−λ > 0. In addition ∞ X X λx e−λ f (x) = x! x=0 x∈X = e−λ = e = ∞ X λx x=0 −λ λ x! e 1. Here we have used the Taylor series for the exponential function. The distribution function is for x ∈ X x X λy e−λ P(X ≤ x) = F (x) = y! y=0 This latter summation does not always have an analytic expression. One can use the distribution function or the PMF to answer questions of interest: • what is the probability that X is 1? • what is the probability that X is positive? For the first question: P(X = 1) = f (1) = F (1) − F (0) = λe−λ . For the second question: P(X ≥ 1) = 1 − P(X = 0) = 1 − f (0) = 1 − e−λ .