Cross Validated Asked by rambalachandran on December 6, 2021
A binomial distribution of $n$ samples and probability of success $p$ is defined as
$ P(k) = binom{n}{k} cdot p^kq^{n-k} $.
For a given value of $r$ where $r in mathbb{N} quad text{and } r gt 1$, if we decrease the probability of success by $hat{p} = frac{p}{r}$ and increase the samples $hat{n} = nr$, I want to prove
$P(kge 1) – P(hat{k}ge 1) gt 0 $
$Rightarrow left(1-P(k= 0)right) – left(1-P(hat{k}= 0)right) gt 0$
$Rightarrow P(hat{k}= 0) – P(k= 0) gt 0$
$Rightarrow binom{nr}{0} cdot hat{p}^0hat{q}^{nr} – binom{n}{0} cdot p^0q^{n} gt 0$
$Rightarrow left(1-hat{p}right)^{nr} – left(1-pright)^{n} gt 0$
$Rightarrow left(1-frac{p}{r}right)^{nr} – left(1-pright)^{n} gt 0$
I’m able to show it for specific examples but unfortunately unable to prove the last inequality in generic form.
First, by expanding $(1-p/r)^r$ to the power series, we have
$(1-p/r)^r=1-p + [(r-1)p^2/(2! r)-(r-1)(r-2)p^3/(3!r^2)+…+(-p/r)^r]$
that is,
$(1-p/r)^r = 1-p + A$
where $A > 0$ is the term in the square brackets. Thus, we have
$(1-p/r)^r > 1-p$, (for r > 1). Then we have
$left(1-frac{p}{r}right)^{nr} gt left(1-pright)^{n}$
Answered by user295357 on December 6, 2021
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