Monotonicity of the CDF of a Binomial Distribution

Question

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

Answer 1

(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 $p$.

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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 $X\le k$).

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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 $F(n+1,k,p)<F(n,k,p)$

Answer 2

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 $n$.

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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

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

Answer 3

One can answer the first question, “How does F(k;n,p) change when $p$ increases, with $k$, $n$, fixed” by rewriting

\begin{equation*} F(k;n,p) = \sum_{i=0}^{\lfloor k\rfloor} \binom{n}{i} p^i (1-p)^{n-i} = 1-P[X \ge k] = 1 - E[1_{\{X \ge k\}}] \end{equation*}

for $X \sim \mathit{Binom}(p,n)$, noticing that the indicator function $1_{\{x \ge k\}}$ is monotonic in $x$, and then using the fact that the expectation of such monotonic functions is monotonic in $p$, which shows that $F(k;n,p)$ decreases with increasing $p$.

Answer 4

I could not figure out why $E[X^k]\le E[X]$ in the accepted answer, somebody might be able to explain this a little bit?
Here an alternative approach to prove $F(k;n,p)$ decreasing with $p$ is provided. It can be proved that $$F(k;n,p)=P(X \ge k)= P(Y\le p)=F_Y(p)$$ where X is $Binomial(n,p)$ distributed and Y is $Beta(k+1,n-k)$ distributed. Clearly $F_Y(p)$ decreases with $p$, so does $F(k;n,p)$.
Of course, we can differentiate $F_Y(p)$ which is given by $$P(Y \le p) = (n-k){n\choose k}\int_{p}^1t^k(1−t)^{n−k}dt$$ with respect to $p$. This gives $$ \frac{dP(Y\le p)}{dp}=-(n-k){n\choose k}p^k(1-p)^{n-k}<0$$ when $p \in (0,1)$.