TITLE: How to find the CDF of the binomial distribution in terms of an integral
QUESTION [6 upvotes]: This wiki page says that the CDF of the binomial distribution in terms of the beta function can be expressed as
$$F(k;n,p)=Pr(X\leq k)=(n-k){{n}\choose{k}}\int_0^{1-p}t^{n-k-1}(1-t)^k {d}t$$
How to prove the equality? I don't quite have the access to the book referenced by the wiki page( Wadsworth, G. P. (1960). Introduction to probability and random variables. USA: McGraw-Hill New York. p. 52.)

REPLY [5 votes]: Using integration by parts we obtain:
$$F(k;n,p)=Pr(X\leq k)=(n-k){{n}\choose{k}}\int_0^{1-p}t^{n-k-1}(1-t)^k {d}t$$
$$=(n-k){{n}\choose{k}}\int_0^{1-p}(1-t)^k {d}\frac{t^{n-k}}{n-k}$$
$$=(n-k){{n}\choose{k}}\frac{1}{n-k}\left[(1-t)^kt^{n-k}|_0^{1-p}-\int_0^{1-p}t^{n-k}d(1-t)^k\right]$$
$$={{n}\choose{k}}\left[(1-1+p)^k(1-p)^{n-k}+k\int_0^{1-p}t^{n-k}(1-t)^{k-1}dt\right]$$
$$={{n}\choose{k}}p^k(1-p)^{n-k}+{{n}\choose{k}}k\int_0^{1-p}t^{n-k}(1-t)^{k-1}dt$$
$$={{n}\choose{k}}p^k(1-p)^{n-k}+{{n}\choose{k}}\frac{k}{n-k+1}\int_0^{1-p}(1-t)^{k-1}dt^{n-k+1}= etc.$$