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This is from Feller's Introduction to Probability Theory and Its Applications. In the context of Bernoulli trials, we define:

$b(k;n,p) = \binom{n}{k}p^kq^{n-k},$ $P\{S_n \ge r\} = \sum_{v=0}^{\infty}b(r+v;n,p).$

The latter being the probability of having at least $r$ successes. Now, supposing $r \gt np$ and knowing that

$\frac{b(k; n,p)}{b(k-1;n,p)}=\frac{(n-k+1)p}{kq}=1+\frac{(n+1)p-k}{kq},$

show that

$P\{S_n \ge r\} \le b(r;n,p)\frac{rq}{r-np}.$

According to Feller, it follows from the obvious fact that the terms of the series decrease faster than the terms of a geometric series with ratio $1-\frac{r-np}{rq}$. However, it's not obvious for me and I don't see how the upper bound follows.

1 Answers 1

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First notice that

For k > r we have that

$\frac{(n-k+1)p}{kq} <\frac{(n-k+1)p}{rq}$ (as $n \ge k-1$)

Now $\frac{(n-k+1)p}{rq} \leq \frac{(n-(r+1)+1)p}{rq} = \frac{(n-r)p}{rq}$ as the numerator is the largest when $k = r+1$

Now $1 - \frac{(n-r)p}{rq} = \frac{rq - np + rp}{rq} = \frac{r -np}{rq}$ (as $p+q=1$).

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    @AR. Yes, you got it right.2010-08-23