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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.