10
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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

5
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.