# Cumulative distribution function of the Binomial Distribution for fixed probability of success is decreasing in the number of trials

Fix $$0 \le p \le 1$$. The cumulative distribution function of the Binomial Distribution $$B(n, p)$$, which counts the number of successes out of $$n$$ independent trials, each of which has probability $$p$$ for success, is $$F_n(k) = \sum_{i\le k} \binom{n}{i} p^i (1-p)^{n-i}.$$

For any fixed $$k\ge 0$$, is $$F_n(k)$$ decreasing in $$n$$, for $$n\ge k$$?

Yes: Model the binomial distribution as the amount of successes in $$n$$ independent coin tosses $$X_1, \dots, X_n$$ with success probability $$p$$. Let $$B_n$$ be the amount of successes in the first $$n$$ tosses. By definition, $$B_n$$ is now $$B(n,p)$$ distributed and $$B_{n-1}$$ is $$B(n-1,p)$$-distributed. \begin{align} F_n(k) &= \mathbf P(B_n≤k) \\ &= \mathbf P(B_{n-1}+X_n ≤ k) \\ &= \mathbf P(B_{n-1}≤ k-X_n) \\ &≤ \mathbf P(B_{n-1} ≤ k) \\ &= F_{n-1}(k) \end{align} So $$F_n(k)$$ is indeed decreasing in $$n$$.
In words: The probability of $$k$$ or less successes in $$n$$ throws must be lower than $$k$$ or less successes in $$(n-1)$$ throws, because the former implies the latter.
• $$\Pr(B_n\leq k)=\frac{n!}{k!(n-k-1)!}\int_p^1t^k(1-t)^{n-k-1}dt$$ is good to know. Commented Dec 17, 2023 at 9:32