Conjecture: Let $m$ and $n$ be fixed positive integers and let $f(k)$ be the probability that a Binomial($k(m+n)$, $p$) random variable is less than $kn$. Then for sufficiently small $p$, $f(k)$ is an increasing function of $k$.

Unless I've made a mistake, the conjecture holds if you replace the binomial with its normal approximation and $p < n/(m+n)$. This suggests the conjecture should be true for large $m$ and $n$. But is it true for all $m$ and $n$, i.e. for small values? How large can $p$ be?

  • $\begingroup$ Just to clarify: in your conjecture, is $f$ supposed to be an increasing function for all sufficiently small $p$ and all $m$ and $n$? (This is not exactly what you stated but it seems to be what you mean) $\endgroup$
    – Yemon Choi
    Jul 4, 2013 at 3:07
  • $\begingroup$ Yes. I'll edit the question to make that clearer. $\endgroup$ Jul 5, 2013 at 11:48
  • $\begingroup$ Two clarifying remarks: 1) I understand it is supposed $p>0$ (i.e. non-degenerate distribution) as with $p=0$, $f(k)$ is obviously a constant function 2) I think that we should require $k$ and $n$ to be positive integers and $m$ to be a non-negative one; e.g. with $k, n, m$ being negative integers $f(k)$ is a decreasing function. $\endgroup$
    – Waldemar
    Jul 8, 2013 at 8:17

1 Answer 1


Unless I've made a mistake: the conjecture is true for $p<kn/(kn+km+1)$ and possibly other values as well.

Requiring $k(m+n)$ to be an integer, $f(k) = \displaystyle\sum_0^{\lfloor kn\rfloor}(km+kn)Cr p^r (1-p)^km+kn-r$ and a similar expression with the same limits gives a lower bound for $f(k +1/(m+n))$. Considering the ratio of equivalent terms in this sum: if this is greater than $1$, $f(k + 1/[m+n])>f(k)$ which happens if $p<kn/(km+kn+1)$. Hope this helps!


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