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Let $p$ and $q$ be probability densities on $\mathbb R$, with respect to the Lebesgue measure $dx$. The corresponding Hellinger integral and distance are $H(p,q):=\int_{\mathbb R}\sqrt{pq}\,dx$ and $\rho(p,q):=\sqrt{\frac12\int_{\mathbb R}(\sqrt{p}-\sqrt q)^2\,dx}=\sqrt{1-H(p,q)}$.

For real $t$, define the $t$-shifted version $p_t$ of $p$ by the formula $p_t(x):=p(x-t)$ for real $x$. A general question is this: Are there broad conditions that guarantee that $H(p_0,p_t)$ will be nonincreasing in $t\ge0$?

A more specific question: if $p$ is unimodal (that is, nondecreasing to the left of some point $c$ and nonincreasing to the right of $c$), will it guarantee that $H(p_0,p_t)$ is nonincreasing in $t\ge0$? This is easy to see if $p$ is also assumed to be symmetric. Numerical experiments (with piecewise-constant $p$) suggest that the unimodality should be enough, even without the symmetry.

(Of course, for (say) saw-like $p$'s, we will not have the desired monotonicity.)

To put this into a context: if one has the strict version of the desired monotonicity, this will allow the convenient reparameterization $[0,\infty)\ni t\mapsto\tau:=\rho(p_0,p_t)$ of the statistical parametric shift (location) family $(p_t)$ of densities.

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  • $\begingroup$ Why don't you simply differentiate w.r.t. $t\geq 0$? $\endgroup$
    – leo monsaingeon
    Commented Jan 5, 2016 at 7:19
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    $\begingroup$ If the probability density is log-concave function then the claim is true. Indeed, $\sqrt{p(x)p(x+t)}$ is log-concave, then by Prekopa--Leindler $f(t)=\int_{\mathbb{R}}\sqrt{p(x)p(x+t)} dx$ is log-concave, and by change of variables $f(t)$ it is even. Therefore $f(t)$ is decreasing for $t\geq 0$. Actually this argument can be used to show that the claim is true if $B(x,y)=p(x)q(y)$ is quasiconcave function. $\endgroup$
    – Paata Ivanishvili
    Commented Jan 5, 2016 at 8:35
  • $\begingroup$ leo: so, what should I do after the differentiation? $\endgroup$
    – Iosif Pinelis
    Commented Jan 5, 2016 at 14:16
  • $\begingroup$ Paata: Your point about the log-concave case is very nice. However, in the problem from which my question came, the log-concave case can be easily tackled without reparameterization. Can you elaborate on the "quasiconcave" part of your comment? In particular, I don't know what you mean by $q$. In my question, I only have $p$ and its shifts $p_t$. $\endgroup$
    – Iosif Pinelis
    Commented Jan 5, 2016 at 14:43
  • $\begingroup$ Iosif, I am sorry I meant $p(x)p(y)$ is quasiconcave. It was a typo in my previous comment. This case does not solve your conjecture for unimodal functions, but at least it covers some cases of it. Yes I will write details later because I am not with my computer. $\endgroup$
    – Paata Ivanishvili
    Commented Jan 5, 2016 at 16:48

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I'm pretty sure that you figured it out by now, but I'll post it for the sake of completeness. Any unimodal $\sqrt{p}$ can be approximated by a sum $\sum c_k\chi_{I_k}$ with $I_1\subset I_2\subset\dots$. Now just notice that the function $\int(\chi_{I_k})(\chi_{I_j})_t$ is non-increasing in $t\ge 0$ for any $k,j$.

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