 Centred interval maximizes mass for an even unimodal density

Let $p$ be a probability density function on $\mathbb{R}$ with an even function representative satisfying $p(-t)=p(t)$, and suppose $p$ is nonincreasing for $t\ge 0$. For every half-width $a\ge 0$ and center $d\in \mathbb{R}$,Thus a centered interval maximizes captured probability among all intervals of the same width. Symmetry alone is insufficient: mass concentrated in two distant peaks can favor an interval around one peak. The single-central-maximum condition supplied by unimodality is essential.

Here is a direct proof, including densities with jumps. If $F$ is the cumulative distribution function, then $M(d)=F(d+a)-F(d-a)$. Integrability of the density gives absolute continuity of $F$, and hence of $M$ on every bounded interval. Its derivative exists almost everywhere and satisfiesFor $d\ge 0$, one has $|d+a|\ge |d-a|$. Evenness and radial monotonicity therefore give $p(d+a)\le p(d-a)$, so $M^{\prime }(d)\le 0$ almost everywhere. Absolute continuity implies that $M$ is nonincreasing on $[0,\infty )$. Symmetry gives $M(-d)=M(d)$, proving the claim for negative centers too. When $a=0$, all integrals vanish.

An original example uses the triangular distribution density $p(t)=1-|t|$ for $|t|\le 1$ and zero elsewhere. At half-width $a=1/2$,Moving the window a quarter-unit away from the center loses probability even though its length stays1.

The maximizer need not be unique. For a uniform distribution on $[-2,2]$ and $a=1/2$, every center $d\in [-3/2,3/2]$ captures probability1/4. A flat density gives a plateau of equally good intervals. The non-strict form of unimodality, described through interval-shaped superlevel sets, is therefore the appropriate general hypothesis.^[ref]

The lemma applies to a detection time gate when the normalized signal profile is an even unimodal density and the objective is captured signal area. Background noise, asymmetric response or timing uncertainty can change the best gate for another objective; maximizing captured area alone does not optimize every measurement criterion.