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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.

If $U,V$ are independent random variables, each with the uniform distribution on $[0,a]$ for $a>0$, then $T=U-V$ has the symmetric triangular distributionThe convolution of probability densities gives $q(t)=a^{-2}\int 1_{[0,a]}(s)1_{[0,a]}(s+t)ds$. The overlap of these two intervals has length $a-|t|$ for $|t|<a$ and is empty otherwise, proving the formula. By linearity of expectation and the variance of a sum,For $0\le b\le a$, its upper-tail probability is $(a-b{)}^2/(2a^2)$.

A useful model for elapsed time measured from two independently truncated clock readings uses $a$ as the display resolution and $T$ as the endpoint-offset difference. Independence is essential: uniform marginal probability distributions alone do not imply this triangular distribution. For example $U=V$ gives $T=0$ with certainty. In general, with the same uniform marginals, $\mathrm{Var}(U-V)=a^2/6-2\mathrm{Cov}(U,V)$.

Use the uniform proposal on $[1,7]$, with density $g=1/6$. The triangular distribution peaks at $x=3$ with height $1/3$, so the optimal constant for this proposal is $M=2$. At each attempt take two fresh pseudo-random uniforms $U,V$, and propose $Y=1+6U$. The rejection method accepts according toIndeed these bounds are $f(Y)/(2g(Y))=3f(Y)$. Part (a) proves that the accepted value has the required density. Acceptance probability is $1/2$, so the method needs two proposal attempts on average, and hence four uniforms on average if each attempt uses two. The validity of the ideal algorithm presumes independent uniform draws; merely belonging to $[0,1]$ does not by itself make a pseudo-random sequence uniform or independent.

The graph is a rectangle of height one over $[0,1)$, with zero height elsewhere. Its unit area is the required normalization of the probability density function.

Let $E$ be the finish-time offset from the start of its displayed minute. The elapsed time is $25+E-S$, so $T=E-S$. To obtain the stated density, use the intended model that $E$ and $S$ are independent random variables, both with the uniform distribution on $[0,1)$.

Their joint probability density function is one on the unit square. Equivalently, the convolution of probability densities givesThe integrand is one exactly where $s\in [0,1]$ and $s+t\in [0,1]$. The overlap has length $1+t$ for $-1\le t<0$, length $1-t$ for $0\le t<1$, and zero otherwise. ThereforeThis is a symmetric triangular distribution. The independence and finish-time uniformity are additional modelling assumptions: the two truncated clock readings alone do not imply them. For example, if $E=S$ then the elapsed time is exactly twenty-five minutes and $T=0$, even though each endpoint offset is individually uniform.