Let m=1/2 and n=1/2 be the probabilities of success and failure respectively and let p(k) be a probability mass function such that p(k)=m \sum_{i=1}^{k}n^i+n \sum_{i=1}^{k} m^i.

Celinamg8 2022-09-17 Answered
Let m = 1 2 and n = 1 2 be the probabilities of success and failure respectively and let p(k) be a probability mass function such that p ( k ) = m i = 1 k n i + n i = 1 k m i .
Is this a valid probability mass function?
Say we carry out this test infinitely many times, does p ( k ) = 1?
We recognise that the probability mass function is a sum of two slightly shifted geometric series. Let's consider one of the two (the other will follow the exact same argument).
Say we are summing from 0 to , that is let
S 0 = i = 0 n i .
Then this is simply evaluated as S 0 = 1 + n + n 2 + n 3 + = 1 1 n
which is perfectly valid as n = 1 2 . If, then, I start this sequence at i = 1, is it valid to say that
S 1 = i = 1 n i = 1 1 n 1 = 1 ( 1 n ) 1 n = n 1 n
and as such that, as k , we have something like
m n 1 n + m n 1 m = 1.
Have I made any ridiculous logical jumps or is this valid?
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Answers (1)

ahem37
Answered 2022-09-18 Author has 14 answers
Step 1
Using your formula we obtain p ( 1 ) = 1 2 and p ( 2 ) = 3 4 . Thus p ( 1 ) + p ( 2 ) > 1, which is impossible.
Step 2
If we define the distribution of a random variable X by saying that Pr ( X k ) is given by your sum, then we will have defined a legitimate distribution. The probability mass function is then given by p ( k ) = m n k + n m k . Since m = n = 1 2 , it would be clearer to say p ( k ) = 1 2 k for every positive integer k.

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