3
$\begingroup$

Are there any well-known conditions that guarantee that a probability distribution isn't too "spiky"?

I ask this question because I am interested in the families of probability distributions $f(x)$ on the unit interval such that the following criterion holds: there exists a measurable subset $S\subset[0,1]$ such that $$4\left(\int_{S}f(x)\,dx\right)^{2}\geq\lambda(S)\int_{0}^{1}f(x)^{2}\,dx$$where $\lambda(S)$ denotes the 1-dim real Lebesgue measure of $S$, i.e. the sum of the intervals that comprise it. This looks like a reversed Jensen's inequality, except for the fact that we're taking integrals over two separate domains.

(1)Is there a well-known sufficient condition that would cause this to hold?

(2)How about if I increase the coefficient $4$?

$\endgroup$
  • $\begingroup$ Could you also add your reference indicating where such a measure comes from? $\endgroup$ – Henry.L Dec 23 '17 at 19:29
1
$\begingroup$

Non-Gaussianness is an ambiguous concept. In the continuum of probability distributions such as the uniform, where all events are clustered into a given range and equally likely. On the other side are structured, spiky distributions with the certain event being the extreme example.[1]

Therefore the measure of spikiness is usually based on estimators of a shape parameter of an assumed data generating distribution. For example, If you assume the data is generated from a beta distribution, then the spikiness can be measured by an estimator of its shape parameter. This is the classic thinking when a parameterized model is assumed for the underlying probability distribution that generates the model.

Following this idea, a classic test of comparing how similar two probability distributions are is the Kolmogorov-Smirnov test. It induces a nonparametric measure of similarity, and therefore could be used for exploring spikiness. In this direction of characterizing spikiness. In other words, spikiness can be measured by an appropriate choice of norm on the space of probability distributions supported on $[0,1]$.

To be honest I think this is more like a reverse Schwarz inequality rather than a Jesn inequality since I do not see how convexity comes into play. If that is the case, then such a sufficient condition reduces to a choice of $S$ such that majorant conditions hold. For any isotonic functional $A$, including most norms, $0\leq A(f^{2})A(g^{2})-A^{2}(fg)\leq\frac{1}{4}(M-m)^{2}A^{2}(g^{2})$ where $m\cdot g\leq f\leq M\cdot g$ In this case we can take $f=g$ and see if we can related the majorant coefficients $M,m$ with the $\lambda(S)$, which I believe is a common pratice in deriving a bound since the above inequality provides a sharp bound.

[1]Gray, William Charles. Variable norm deconvolution. No. 19. Ph. D. thesis: Stanford University, 1979.

[2]Dragomir, Sever S. "Reverses of Schwarz inequality in inner product spaces with applications." Mathematische Nachrichten 288.7 (2015): 730-742.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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