Suppose $X \sim Binomial(n,p)$ with known $n$ but unknown $p$, and let $G(p,k) = P[X \geq k)$ for $k=0, \ldots, n$. I am looking for a tight upper bound on $|G(p_1, k) - G(p_2, k)|$ for some given $k$.

Equivalently, for a given $k$, we know that $G(p,k)$ is the regularized incomplete beta function, i.e., $G(p,k) = \frac{1}{B(k,n-k+1)} \int_{0}^p t^{k-1}(1-t)^{n-k} dt $. Therefore, assuming $p_2 \geq p_1$ without loss of generality, $|G(p_1, k) - G(p_2, k)| = \frac{1}{B(k,n-k+1)} \int_{p_1}^{p_2} t^{k-1}(1-t)^{n-k} dt$.

What is a tight upper bound for the last expression?

Can we say, for example in the spirit of Lipschitz continuity, that $|G(p_1, k) - G(p_2, k)| \le C |p_1 - p_2|$ for some constant $C$ which is independent of $n$ and $k$?

  • 1
    $\begingroup$ What is the problem with using the correct, fully accurate value for the difference? The point is that asking for an "upper bound" is meaningless until you specify what kind of function would qualify as an answer. Do you really mean to limit the functions to constant multiples of $|p_1-p_2|$? $\endgroup$ – whuber Apr 13 '18 at 15:05
  • $\begingroup$ @whuber "the problem with the correct, fully accurate value for the difference" is that it depends on values of $n, k, p_1$ and $p_2$. As an illustration, consider the (statistical) case where we observe $X_1, \ldots, X_N \sim Bin(n,p)$. As before $p$ is unknown. Then $\hat{p} = p + O_P(1/N)$, i.e., using $p_1 = p$ and $p_2 = \hat{p}$ we know that $|p_1 - p_2| = O_P(1/N)$ even though we don't know what $p_1$ is. An upper bound that depends on $|p_1 - p_2|$ but not on $p_1$ and $p_2$ will be useful here. Similar examples can be constructed to show the usefulness of a bound free from $n, k$. $\endgroup$ – lordcretin Apr 13 '18 at 15:28
  • $\begingroup$ The confusing aspect of this is that any bound that does not depend on $n$ or $k$ and only on $|p_1-p_2|$ will have to be so crude as to be worthless for any conceivable application. $\endgroup$ – whuber Apr 13 '18 at 15:29
  • $\begingroup$ It does appear non-trivial. I can prove trivially, for example, that $|G(p_1,k) - G(p_2,k)| \le n|p_1 - p_2|$ but that is not a tight enough bound. $\endgroup$ – lordcretin Apr 13 '18 at 15:38
  • $\begingroup$ But since $n$ is arbitrarily large and you require (as stated in the last sentence of the question) that the bound not depend on $n,$ that's an utterly trivial bound! $\endgroup$ – whuber Apr 13 '18 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.