The binomial PMF (probability of exactly k successes in n trials with probability p)
$$f\left( {k,n,p} \right) = {{n!} \over {k!\left( {n – k} \right)!}}{p^k}{\left( {1 – p} \right)^{n – k}}$$
And the recurrence relation for an additional success is
$$f\left( {k + 1,n,p} \right) = {{n – k} \over {k + 1}}\;{p \over {1 – p}}f\left( {k,n,p} \right)$$
The CDF (probability of at most k successes) is
$$F\left( {k,n,p} \right) = \sum\limits_{i = 0}^k {{{n!} \over {i!\left( {n – i} \right)!}}{p^i}{{\left( {1 – p} \right)}^{n – i}}} $$
My question is, is there a similarly simple recurrence relation for the CDF for an additional trial?
$$F\left( {k,n + 1,p} \right) = ??$$
Thank you.
Best Answer
$$\binom{n+1}{i} = \binom{n}{i} + \binom{n}{i-1}$$
$$ \begin{align} F(k,n+1,p) &= \sum_{i=0}^{k}\binom{n+1}{i}p^{i}(1-p)^{n+1-i}\\ &= \sum_{i=0}^{k}\binom{n}{i}p^{i}(1-p)^{n+1-i} + \sum_{i=1}^{k}\binom{n}{i-1}p^{i}(1-p)^{n+1-i}\\ &= (1-p)F(k,n,p) + p\sum_{j=0}^{k-1}\binom{n}{j}p^j(1-p)^{n-j}\\ &= (1-p)F(k,n,p) + pF(k-1,n,p) \end{align} $$
I think you also need this, $$F(k-1,n,p) + \binom{n}{k}p^k(1-p)^{n-k} = F(k,n,p)$$