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For this stellar population, let $m=M/M_{\odot }$ denote initial mass in units of the solar mass, and let $\tau =t/\mathrm{Gyr}$ denote present stellar age. Constant formation of equal numbers of stars per unit time makes the stellar age a uniform distribution on $[0,10]$ Gyr, so $Y=\tau /10$ is uniform on $[0,1]$. The normalized initial mass function has probability density function $f(m)=0.1m^{-2}$ for $m>0.1$, since ${\int }_{0.1}^{\infty }{m}^{-2}dm=10$. ConsequentlyThese formulae have the stated age and mass domains; outside them the relevant cumulative fractions saturate at zero or one. The probability integral transform makes $X$ uniform on $[0,1]$: $m=0.1/X$ gives $f(m)|dm/dX|=1$. The time-independent initial mass function and constant number formation rate give independence of $X$ and $Y$. All fractions here count objects, including white dwarfs, using the stipulated stellar evolution law.

A red giant hasThus the mass boundaries are $m=9/\tau$ and $m=10/\tau$ for positive $\tau$. Equivalently, for $0.9<m<1$ the red giant region runs from $9/m$ to the age cap $10$, and for $m\ge 1$ it runs from $9/m$ to $10/m$. There are no red giants with $m\le 0.9$. Boundaries have zero probability and their endpoint convention does not affect the fractions. In the uniform $(X,Y)$ square the red giant region is $9X<Y<10X$, with triangle vertices $(0,0)$, $(1/10,1)$ and $(1/9,1)$. The white dwarf region is the triangle $0<X<Y/10$.

At fixed $Y=y$, define the conditional probabilities of a red giant, white dwarf and main sequence star by $g(y)$, $w(y)$ and $s(y)$. Their interval widths areIntegration over the uniform distribution of age gives the individual-star fractionsThe systems form a coeval binary population: the two binary star components have the same age. Their masses are independent, so $(X_1,X_2,Y)$ has uniform probability density function on $[0,1{]}^3$ and their states have conditional independence given $Y$. Unconditional independence of their states would be incorrect: older systems make both evolved states more likely. The law of total probability now givesThe subtraction removes the double counting of systems containing two red giants.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

For the planar uniform distribution with a Lebesgue measure density, assume that the compact convex set $\mathcal{C}$ has nonempty interior, equivalently positive area. This assumption is missing from the wording: a line segment or singleton would not have such a planar uniform density.

The hit-and-run sampler at a current point $x$ chooses $u=(\cos \Theta ,\sin \Theta )$ with $\Theta$ uniform on $[0,2\pi )$, finds the intervaland samples $T$ uniformly on that interval, setting $x^{\prime }=x+Tu$. Convexity makes the intersection an interval and compactness makes it bounded.

If $x$ is in the interior, every direction has positive chord length, and the next point is in the interior almost surely. For a definition valid at every boundary point as well, redraw the direction whenever the chord has zero length. We use this convention in the following parts. Positive-length directions have positive angular probability at every $x$, since an interior ball is visible from $x$, so the redraw terminates almost surely.

The Markov kernel has the uniform distribution on $\mathcal{C}$ as its invariant law, as follows from the density symmetry proved next. Repeating the step therefore produces approximate uniform samples after convergence. The boundary convention matters: if singleton chords were instead treated as holding moves, a square's corner would have a positive atom at its starting point, contradicting part (b)'s claim of an everywhere absolutely continuous kernel.

Write $\ell (x,u)=t_{+}(x,u)-t_{-}(x,u)$ and letThis is the accepted fraction of directions when zero-length chords are redrawn; it equals one in the interior. The Markov kernel integrates a test function $h$ asThe signed polar representation has Jacobian determinant $|t|$. Each point $y\ne x$ is represented twice, once with positive $t$ and once with negative $t$ and the opposite direction. Since the chord is the same in either representation, the hit-and-run kernel density isfor almost every $y\in \mathcal{C}$, and it is zero outside the body. Thus the transition is absolutely continuous with respect to planar Lebesgue measure. Redrawing removes the only potential zero-chord atoms.

Let $D=\mathrm{diam}(\mathcal{C})>0$. For distinct $x,y\in \mathcal{C}$, the chord contains their segment, so it has positive length, andHence $p(x,y)\ge 1/(\pi D^2)$. Values on the diagonal, and on other null sets where a density is only specified almost everywhere, can be assigned this same lower bound. This gives a version satisfyingFor interior $x,y$, $a(x)=a(y)=1$ and the chord length is unchanged on interchanging its endpoints, so $p(x,y)=p(y,x)$. A convex body's boundary has area zero; for example contractions about an interior point fill the interior and their areas tend to the body's area. Thus this symmetry holds almost everywhere for two uniform points, proving detailed balance and stationarity of the uniform distribution. Without positive area or the boundary convention in part (a), the literal assertion need not hold.

Let $A=\mathrm{Leb}(\mathcal{C})>0$ and let $\mu$ be the uniform distribution on the body. The hit-and-run kernel density lower bound impliesFor the last range, $\mathcal{C}$ lies in a radius-$D$ disc about any of its points, so $A\le \pi D^2$ and $\epsilon \le 1/2$. This global Doeblin condition proves irreducibility with respect to $\mu$ and aperiodicity. The entire state space is a small set; $V=1$, $\lambda =1/2$, and $b=1/2$ satisfy the geometric drift condition.

In fact the conclusion is stronger. Write $P=\epsilon \Pi +(1-\epsilon )R$, where $\Pi (x,\cdot )=\mu$ and $R$ is the residual Markov kernel. Since $\mu P=\mu$, also $\mu R=\mu$. Each transition has a probability $\epsilon$ of resetting to stationarity, and after such a reset every subsequent distribution remains stationary. Equivalently,Thus the algorithm has uniform geometric ergodicity, using the full-dimensional and boundary conventions stated in part (a).

Let $S={\sum }_{i=1}^n{X}_i$ be the number of heads. The likelihood function for the Bernoulli distribution sample is proportional to ${\theta }^S(1-\theta {)}^{n-S}$. A uniform distribution prior on $[0,1]$ is the $\mathrm{Beta}(1,1)$ distribution, so Beta-binomial conjugacy gives the posterior distribution

For every admissible transport map $T$, the pushforward measure condition gives $T(x)\in [1,2]$ for $\mu$-almost every $x\in [0,1]$. Hence $|T(x)-x|=T(x)-x$, andApply the Jensen inequality to the convex function $h$ and the probability measure $\mu$:Translation by one sends the source uniform distribution to the target uniform distribution, so $T^{†}(x)=x+1$ is admissible. Its displacement is constantly one, givingNo monotonicity assumption on $h$ is needed, because all displacements have the same sign. The same argument with a transport plan also gives the identical minimum for the Kantorovich optimal transport problem.

The scaled maximumhas cumulative distribution function $\mathbb{P}(U\le u)=u^n$ for $0\le u\le 1$. ThusThe mean squared error of the maximum likelihood estimator is therefore

Each $X_i$ from the uniform distribution on $[0,2\theta ]$ has mean $\theta$ and variance ${\theta }^2/3$. The sample mean is unbiased and has variance ${\theta }^2/(3n)$, hence

Let $U$ have the uniform distribution on $[0,1]$, and setThe variables have the same law. They are not independent random variables, since

A cipher has perfect secrecy when the conditional probability of every plaintext given a ciphertext equals its prior probability: $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$ whenever $\mathrm{Pr}(C=c)>0$. Equivalently, the plaintext and ciphertext are independent random variables.

Fix a ciphertext $c$ that can occur. Perfect secrecy requires every possible plaintext $m$ to remain possible after observing $c$. For a deterministic decryption rule, two different plaintexts producing $c$ require two different keys, so the key space has cardinality at least that of the message space.

In a one-time pad, the message and key belong to the same finite additive group, the key is uniform and independent of the message, and encryption is $C=M+K$. For every $m,c$ there is exactly one key $k=c-m$, soindependent of $m$. Bayes theorem then gives $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$, proving perfect secrecy.

Uniform orientation on the unit sphere means that probability is proportional to the solid-angle elementBecause the minor axis is unoriented, it is enough to take $0\le i\le \pi /2$. After integration over the azimuth and normalization,Equivalently, the random variable $\mu =\cos i$ has the uniform distribution on $[0,1]$.

For the uniform distribution $\mathcal{P}(q)=1$, substitute $q=Q\sin \theta$ into part d:where $K$ is the complete elliptic integral of the first kind. Both $Q$ and $K(Q)$ increase on $(0,1)$; indeed $\mathcal{P}(Q)\sim \pi Q/2$ near zero and diverges logarithmically as $Q\uparrow 1$. Thus the apparent distribution is strongly skewed toward higher $Q$: projection makes many intrinsically flattened systems look round.

Conditionally on $X_1^n=x_1^n$, the uniform index $J$ selects the symbol $a$ with probability ${\hat{P}}_{x_1^n}(a)$. The law of total probability therefore givesThus $\overline{P}$ is the conditional mean type of a string in $B$.

For $0<j_1<\cdots <j_n<t$, the probability of one jump in each infinitesimal interval around $j_1,\dots ,j_n$ and no other jump by time $t$ gives the joint densityfor $(J_1,ldots,J_n)$ together with the event $N_t=n$. Sincethe conditional density ison that ordered simplex.

Now $n$ independent uniform points on $[0,t]$ have joint density $t^{-n}$ on the cube. Each ordered vector has $n!$ permutations, so their order statistics have density $n!/t^n$ on the same simplex. The densities agree, proving Poisson process conditional arrival times.

A one-time pad over the finite additive group ${\mathbb{Z}}_n$ uses a key $K$ that is uniform on ${\mathbb{Z}}_n$, independent of the plaintext $X$, as long as the plaintext, and never reused. Encryption and decryption are

For independent random variables $X,Y$, conditioning reduces entropy and translation by a known element of the finite additive group preserves entropy, soInterchanging $X$ and $Y$ gives $H(X+Y)\ge H(Y)$, henceThis is the entropy of a sum of independent finite-group variables.

The result explains why adding independent pad symbols cannot reduce uncertainty. For a uniform pad, $X+K$ is itself uniform and is independent of $X$, which is the perfect-secrecy property. Independence is necessary: if $X$ is nonconstant and $Y=-X(\mathrm{mod}n)$, then $X+Y=0$ is constant, so

Write the molecular copy numbers as $\mathbf{x}=(x_1,x_2,x_3,x_4)$ and use the power-law reaction propensity convention of the paper. The seven propensities and stoichiometric vectors areA channel is assigned zero propensity whenever its update would leave the nonnegative integer lattice.

The Gillespie algorithm starts from $\mathbf{x}=(5,0,0,0)$ and $t=0$. At the current state compute $a_0={\sum }_{r=1}^7{a}_r$. If $a_0=0$, terminate the path. Otherwise draw independent random variables $U_1,U_2$ from the uniform distribution on $(0,1)$, set the next waiting time toand choose the least index $r$ satisfyingThen update $t\leftarrow t+\tau$ and $\mathbf{x}\leftarrow \mathbf{x}+{\nu }_r$, and repeat. The minimum of the seven competing reaction clocks has an exponential distribution of rate $a_0$, while reaction $r$ wins with probability $a_r/a_0$.