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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(k\ge 1) - P(\hat{k}\ge 1) \gt 0 $

$\Rightarrow \left(1-P(k= 0)\right) - \left(1-P(\hat{k}= 0)\right) \gt 0$

$\Rightarrow P(\hat{k}= 0) - P(k= 0) \gt 0$

$\Rightarrow \binom{nr}{0} \cdot \hat{p}^0\hat{q}^{nr} - \binom{n}{0} \cdot p^0q^{n} \gt 0$

$\Rightarrow \left(1-\hat{p}\right)^{nr} - \left(1-p\right)^{n} \gt 0$

$\Rightarrow \left(1-\frac{p}{r}\right)^{nr} - \left(1-p\right)^{n} \gt 0$

I'm able to show it for specific examples but unfortunately unable to prove the last inequality in generic form.

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1 Answer 1

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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

$\left(1-\frac{p}{r}\right)^{nr} \gt \left(1-p\right)^{n}$

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