Can we generalise the Binomial distribution for a non-integer number of Bernoulli trials?
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asked Apr 20, 2016 at 12:46
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1$\begingroup$ How do you imagine a 4.5th trial for example? $\endgroup$– AlexApr 20, 2016 at 12:48
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$\begingroup$ I am not entirely confident in my answer, since I am still wrapping my head around this thing. Let's say that the binomial will give us the value of X where $X+c < 11$. Now, assume that we know that c = 5.5 $\endgroup$– Dionysios GeorgiadisApr 20, 2016 at 12:58
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1 Answer
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I see at least two (even three) ways to answer your question:
Negatively : There exists a concept of infinitely divisible distribution https://en.wikipedia.org/wiki/Infinite_divisibility_%28probability%29 ; alas, binomial distribution do not belong to this category.
Positively : one can see the issue in a larger perspective :
a) either by considering the natural generalization of (discrete) Binomial Distribution B(n,p) into (continuous) Normal distribution, with pdf $\dfrac{1}{\sigma \sqrt{2 \pi}}e^{-(x-np)^2/(2 \sigma^2)}$ with $\sigma^2=npq$ and taking $n=9/2$ for example.
b) or, if one desires to stay with discrete random variables, define (artificially!!!) $X \sim Bin(4.5,p)$, by taking $X=Y/2$ with $Y \sim Bin(9,p)$.
answered Apr 20, 2016 at 13:15
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$\begingroup$ I see, you really killed it with the infinite divisibility concept. There is not answer to my problem looks like. Out of curiosity, how would you handle the following. Let's say that the binomial will give us the value of X where $X+c<11$. Now, assume that we know that c = 5.5 $\endgroup$ Apr 20, 2016 at 13:32
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$\begingroup$ I have some difficulty to understand your problem because your last sentence is incomplete. $\endgroup$ Apr 20, 2016 at 13:50
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$\begingroup$ Forgive me but my question seems stupid now. I do not believe that it will help anyone if I post it. $\endgroup$ Apr 20, 2016 at 13:57
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$\begingroup$ May I ask you to check (green check in the margin) the answer as validated: it helps people to see if the question has been answered satisfactorily for the asker. Thanks. $\endgroup$ Apr 20, 2016 at 14:00