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Binomial distribution is the distribution of a total number of successes in a given number of Bernoulli trials. The common notation is b(k; n, p), where k is the number of successes, n is the number of trials, p is the probability of success. We know that b(k; n, p) = C(n, k) pk(1 - p)n - k.
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For a fixed number of trials n, the binomial distribution always behaves in the same way: as a function of k, it is monotone increasing up to a certain point m after which (perhaps with an exception of the next point) it is monotone decreasing.
Indeed,
| | b(k; n, p) / b(k-1; n, p) | = C(n, k) pk(1 - p)n - k / C(n, k-1) pk-1(1 - p)n - k + 1 |
| (1) | | = (n - k + 1) p / k(1 - p) |
| | | = 1 + ((n + 1) p - k) / k(1 - p). |
It is clear now that the right hand side in (1) is greater than 1 whenever k > (n + 1) p and it is less than 1 when k < (n + 1) p. It may happen, of course, that m = (n + 1) p is an integer in which case
| | b(m; n, p) = b(m - 1; n, p). |
In any event, there is only one integer m that satisfies
| (2) | (n + 1) p - 1 < m ≤ (n + 1) p. |
Summing up, as a function of k, the expression b(k; n, p) is monotone increasing for k < m and monotone decreasing for k > m, with the exception of one case where (n + 1) p is an integer. In this case, there are two maximum values for m = (n + 1) p and m - 1.
The number m that satisfies (2) is known as the most probable (most likely) number of successes in n Bernoulli trials. As a matter of fact, b(m; n, p) is quite small for a large n, even for a reasonable value of p. It is also always different from the average number of successes. The latter could be found the following way. Letting q = 1 - p, we get
| | ∑ k b(k; n, p) | = ∑ k C(n, k) pkqn - k |
| | | = pqn-1∑ k C(n, k) (p/q)k - 1 |
| | | = pqn-1 n (1 + p/q)n - 1 |
| | | = pqn-1 n (q + p)n - 1 / qn - 1 |
| | | = p n, |
where we used the identity
| | n (1 + x)n - 1 | = ∑ k C(n, k) xk - 1. |
The result for the expected value np = ∑ k b(k; n, p) might have been anticipated given the interpretation of the probability as a relative frequency.
Note that the expected value np is always different from the most likely value (n + 1) p, provided of course p ≠ 0.
Copyright © 1996-2008 Alexander Bogomolny
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