Show that a single binomial probability can be written as the sum of two binomial probabilities

Dulce Cantrell

Dulce Cantrell

Answered question

2022-09-08

Show that a single binomial probability can be written as the sum of two binomial probabilities
I am trying to prove that the sum of two independent binomial probabilities with the same parameter (probability of success in a single trial) is equivalent to a binomial probability with the same parameter , but an index (number of trials) and number of successes that is equivalent to the sum of the indexes from the first two binomial probabilities and the sum of the number of successes from the first two binomial probabilities.
I think the basic idea is that the binomial probability of an event can be broken up and calculated in two "chunks" which can be summed to find the desired binomial probability.
Sorry for not including any mathematical operators in this post; I am unfamiliar with how they are actually written in this type of a post.
Edit: ( n x ) π x ( 1 π ) n x = ( a y ) π y ( 1 π ) a y + ( b z ) π z ( 1 π ) b z where a + b = n and y + z = x

Answer & Explanation

Gerardo Kent

Gerardo Kent

Beginner2022-09-09Added 12 answers

Step 1
Let X be the number of successes when an experiment is repeated independently m times, with probability of success p each time. Now suppose the experiment is done again n times.
Then X has binomial distribution, parameters m and p, while Y is binomial with parameters n and p.
It is clear that X + Y is the number of successes when the experiment is repeated independently m + n times. So X + Y has binomial distribution with parameters m + n and p.
Step 2
Remark: One can also do the problem the hard way. Let W = X + Y. Then
Pr ( W = w ) = i = 0 w Pr ( X = i ) Pr ( Y = w i ) .
Using the usual binomial distribution formulas, we get
Pr ( W = w ) = i = 0 w ( m i ) p i ( 1 p ) w i ( n w i ) p w i ( 1 p ) n ( w i ) .
Manipulation of the above formula yields, after a while,
Pr ( W = w ) = ( m + n w ) p w ( 1 p ) m + n w .

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