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Put $\mu =\mathbb{E}[X]$ and ${\sigma }^2=\mathrm{Var}(X)$. Independence of $X$ and the uniform distribution variable $U$ givesAlso $\mathbb{E}[U^2]=1/3$, soTherefore

An orbit is unbound precisely when its specific orbital energy is nonnegative, equivalently $a/a^{\prime }\le 0$. At periapsis, $f=0$, and for $e=1/2$ the condition from part (d) isThe directions are sampled from the uniform distribution on a circle. The fraction satisfying $\cos \varphi \ge x$ is $\arccos x/\pi$; setting it equal to $1/6$ gives $x=\sqrt{3}/2$. ThereforeThe positive root is

If $X$ has a continuous cumulative distribution function $F$, then $F(X)$ has the uniform distribution on $(0,1)$. Combined with inverse transform sampling, this constructs the monotone rearrangement $G^{-1}(F(X))$ to a target with quantile function $G^{-1}$.

At unit scale the probability density function is $r\exp (-r^2/2)$ for $r>0$. Its square has an exponential distribution of rate $1/2$. Thus $\sqrt{-2\log U}$ has this distribution for a uniform distribution variable $U$ on $(0,1)$.

For positive mixture weights $w_j$ summing to one, set $C_j={\sum }_{i\le j}{w}_i$ and choose $J=j$ when $C_{j-1}\le U<C_j$, where $U$ has a uniform distribution. Then $V=(U-C_{J-1})/w_J$ has a uniform distribution independently of $J$: for $0\le v\le 1$, $\mathrm{Pr}(J=j,V\le v)=w_{j}v$. This permits exact finite mixture model sampling with a recycled continuous random variable.

The square of a uniform distribution variable on $(0,1)$ has cumulative distribution function $\sqrt{x}$ and an integrable inverse-square-root probability density function. Reflection replaces $x$ by $1-x$ in that density.

The stellar age is the elapsed time since a star formed. A constant number formation rate on a finite interval gives a uniform distribution of present ages if every formed object remains counted, including stellar remnants. An age distribution inferred only from surviving main sequence stars generally differs because their lifetimes depend on mass.

For the random adjacent transposition shuffle on $S_n$ with $n\ge 2$, the position $f(\sigma )={\sigma }^{-1}(i)$ of one label is uniform on $\{1,\dots ,n\}$ in equilibrium, so ${\mathrm{Var}}_{\pi }(f)=(n^2-1)/12$. Each adjacent transposition moves the label with probability $2/n$, giving $\mathcal{E}(f,f)=(n-1)/n^2$. The Poincare inequality for a reversible Markov chain variational formula therefore gives $\gamma \le 12/[n^2(n+1)]$. In particular the continuous uniform distribution on $(0,1)$ has variance $1/12$, not $1/6$.

With independent uniform distributions on $[0,1]$ as sequential offers and a finite number of rounds, the continuation value with $j-1$ offers remaining is $v_{j-1}$. Accept the current offer when it reaches that threshold. Integrating the maximum of the offer and continuation value gives the recursion, by the Snell envelope principle.

A Pareto distribution prior for the upper limit of independent uniform distributions remains Pareto after observing data. The likelihood adds the sample size to the shape and replaces the lower bound by the larger of the prior bound and sample maximum.

For positive observations and $m=\max (\beta ,{\max }_j{y}_j)$, the Bayesian model evidence follows by integrating the uniform distribution likelihood against a Pareto distribution prior. For independent groups with shared fixed hyperparameters, these evidences multiply. The resulting hyperparameter log likelihood is $I\log \alpha -{\sum }_i\log (\alpha +n_i)-\alpha {\sum }_i\log (m_i/\beta )$ up to a constant.

A uniform random variable has a uniform distribution. Independent such variables can supply independent inverse transform sampling; reusing a single variable for two transforms generally creates dependence.

For independent uniform distribution increments on $(0,1)$ and $0\le x\le 1$, let $N(x)$ be the first index at which their partial sum exceeds $x$. First-step analysis gives $M(x)=\mathbb{E}N(x)=1+{\int }_0^{x}M(s)ds$, hence $M=e^x$. Equivalently, $\mathbb{P}(N(x)>n)=x^n/n!$ is a simplex-volume probability and the tail-sum formula sums to $e^x$. The strict inequality gives $N(0)=1$ almost surely.