# Monotonicity of the CDF of a Binomial Distribution

Consider the following CDF of a binomial distribution with $$p\in(0,1)$$ and $${\lfloor k\rfloor}\in [0,n]$$ $$\begin{equation*} F(k;n,p) = \sum_{i=0}^{\lfloor k\rfloor} \binom{n}{i} p^i (1-p)^{n-i} \end{equation*}$$

How does $$F(k;n,p)$$ change in the following circumstances

• $$p$$ increase with $$k,n$$ fixed
• $$n$$ increase with $$k,p$$ fixed

and how to prove these monotonicity conclusions?

Some numerical simulation results: $$F(1,3,0.3)=0.784$$, $$F(1,3,0.1)=0.9720$$, $$F(1,4,0.1)=0.9477.$$

• Intuitively: $F(k;n,p)$ is essentially $P(X\le k)$, where $X\sim Binomial(n,p)$. That is, it is the probability that "the number of successes in $n$ Bernoulli trials is small" (where "small" means $\le k$). So if $p$ is increased (and all else is fixed), I would expect this probability to decrease, since increasing $p$ increases the success probability of each trial, so you would expect more successes. Similarly, if $n$ is increased (all else fixed), I would expect more successes to occur, since there are more trials (more opportunities for success), so $F(k;n,p)$ should decrease. – Minus One-Twelfth Aug 10 at 1:20

(Edited)

For the dependence on $$p$$, notice that $$\frac{\partial \, p^i (1-p)^{n-i}}{\partial p}= \frac{i}{p(1-p)}p^i (1-p)^{n-i} -\frac{n}{1-p} \, p^i (1-p)^{n-i} \tag 1$$

Also, if $$X$$ is a $$(n,p)$$ Binomal (hence $$E[X]=np$$), let $$X^{(k)}$$ be $$X$$ truncated to $$[0, \lfloor k \rfloor]$$

Then $$P(X^{(k)}=i)=\frac{1}{F(k,n,p)} \binom{n}{i} p^i (1-p)^{n-i} \,[0\le i \le\lfloor k \rfloor] \tag 2$$ and

\begin{align}\frac{\partial \, F(k,n,p)}{\partial p} &= \frac{1}{p(1-p) }F(k,n,p) E[X^{(k)}]-\frac{n}{1-p} F(k,n,p) \\ &= \frac{F(k,n,p)}{p(1-p)} ( E[X^{(k)}] - E[X] ) \tag 3 \end{align}
But $$E[X^{(k)}] < E[X]$$ (except for the trivial case $$\lfloor k \rfloor = n$$). Hence the derivative is negative and $$F(k,n,p)$$ decreases with $$p$$.

Corrected: I had the wrong sign in $$(3)$$ (confirmation bias!). Now it's correct (checked numerically). And, yes, $$(2)$$ is right, it's the distribution of a truncated Binomial (which of course corresponds to the distribution of $$X$$ conditioned on $$X\le k$$).

Added: For the dependence on $$n$$: letting $$X_{n,p}$$ be a Binomial $$(n,p)$$, and with $$k$$ integer, we have

$$F(n,k,p)= P(X_{n,p}\le k)= \sum_{i=0}^k \binom{n}{i}p^i(1-p)^{n-i}$$

and $$F(n,k-1,p) = F(n,k,p) - \binom{n}{k}p^k(1-p)^{n-k}$$ Hence \begin{align} F(n+1,k,p)&= P(X_{n+1,p} \le k)\\ &= P(X_{n+1,p} \le k \mid X_{n,p} < k) P(X_{n,p} < k) + P(X_{n+1,p} \le k \mid X_{n,p} = k) P(X_{n,p} = k) \\ &= 1 \times F(n,k-1,p) + (1-p) \binom{n}{k}p^k(1-p)^{n-k}\\ &= F(n,k,p) - \binom{n}{k}p^k(1-p)^{n-k} + (1-p) \binom{n}{k}p^k(1-p)^{n-k}\\ &= F(n,k,p) - \binom{n}{k}p^{k+1}(1-p)^{n-k} \end{align} Then $$F(n+1,k,p)

• I think, from simulations, both of them are actually decreasing functions. For fixed k,p while varying n, you also need to consider the effect from the polynomial $p^i(1−p)^{n−i}$ which is decreasing as n increases. Example: $F(1,3,0.3)=0.784$, $F(1,3,0.1)=0.9720$, $F(1,4,0.1)=0.9477$. – Mathexx Aug 7 at 3:05
• In (2), it does not look like a truncated binomial distribution but a conditional probability. Hence, we might not have $E[X^{(k)}]<E[X]$. Also, please check the simulation result in the last comment. – Mathexx Aug 10 at 0:29
• You are right in $F$ decreasing with $p$, Sign fixed. But the rest (including the truncated mean) is right. – leonbloy Aug 10 at 1:48
• You are right. Do you have any hints on the second case: $k,p$ fixed but $n$ varies? From the simulation, $F$ is a decreasing on $n$ does not seem to depend on the selection of $k,p$. – Mathexx Aug 10 at 21:36

Restrict the problem to $$k\geq 0$$ since when $$k<0$$ $$F(k;n,p)$$ is the sujm of binomials with negative bottom numbers, that is, zero.

Let me give heuristic arguments that provide ideas that can easily be made into proofs.

Consider a random variate $$P$$ which is the sum of $$n$$ independent Bernoulli trials with success probability $$p$$. Then $$F(k;n,p)$$ is the probability that $$P \leq k$$. And for $$\Delta p >0$$, $$F(k;n,p+\Delta p)$$ is the probability that $$P \leq k$$ for a sum involving the slightly higher Bernoulli value $$p+\Delta p$$.

Now picture each trial to be done by taking a uniform $$X_i$$ on $$(0,1)$$ and accepting if it is less than $$p$$. For every vector $$\vec{X}$$ of trial outcomes going into $$F(k;n,p)$$ there is a matching vector of $$\vec{X}$$ of trial outcomes going into $$F(k;n,p+\Delta p)$$. And there are also some outcomes contributing to $$F(k;n,p+\Delta p)$$ that were not in $$F(k;n,p)$$, namely, those cases where for some $$i$$, $$p < X_i \leq p+\Delta p$$.

So $$F(k;n,p+\Delta p) > F(k;n,p)$$ and $$F$$ is monotonic increasing with $$p$$.

Now fix $$k$$ and $$p$$ and let $$n$$ increase to $$n+1$$. Well, for all $$i$$,

$$\binom{n+1}{i} = \frac{n+1}{n+1-i} \binom{n}{i} > \binom{n}{i}$$ so each term in the sum is monotonic increasing, so the sum is also monotic increasing with $$n$$.

CORRECTION

$$\binom{n+1}{i} p^{i+1} (1-p)^{n-{i+1}} = \frac{n+1}{n+1-i} \frac{p}{1-p} \binom{n}{i} p^i(1-p)^{n-i}$$ If $$p\geq \frac12$$ this is always greater than $$\binom{n}{i}$$ so each term in the sum is monotonic increasing, so the sum is also monotonic increasing with $$n$$.

If $$p\geq \frac12$$ then this is monotonic decreasing until $$i$$ is large enough that $$(n+1) p > (n+1-i) (1-p)$$. After that, in changes to monotonic decreasing.

The change point comes at

$$i > (n+1) \frac{1-2p}{1-p}$$

• I think, from simulations, both of them are actually decreasing functions. For fixed $k,p$ while varying $n$, you also need to consider the effect from the polynomial $p^i(1-p)^{n-i}$ which is decreasing as $n$ increases. – Mathexx Aug 5 at 19:34
• In the first equation of the CORRECTION, should it be $\binom{n+1}{i} p^{i} (1-p)^{n+1-i}$ instead? – Mathexx Aug 9 at 23:25
• The problem with the "change point" is that it only applies to the terms of the sum, but it doesn't say much about the whole sum. – leonbloy Aug 11 at 17:05