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Let $A,B$ be indicator random variables with conditional independence given a latent variable $T$, and suppose $\mathbb{E}[A\mid T]=\mathbb{E}[B\mid T]=g(T)$. ThenFor a short proof, conditional independence gives $\mathbb{E}[AB\mid T]=g(T{)}^2$. The law of total expectation gives $\mathbb{E}[AB]=\mathbb{E}[g(T{)}^2]$ and $\mathbb{E}[A]=\mathbb{E}[B]=\mathbb{E}[g(T)]$. Subtract their product to obtain the claimed variance. Since $0\le g(T)\le 1$, all the required expected values exist.

In the two-environment example of conditional independence, $g(T)$ is $0.1$ or $0.9$ with equal probabilities. Hence $\mathbb{E}[g(T)]=0.5$ and $\mathrm{Var}(g(T))=((0.1-0.5{)}^2+(0.9-0.5{)}^2)/2=0.16$. The two trials have zero conditional covariance in either environment, yet their unconditional covariance is positive because their success probabilities rise and fall together.

More generally, square-integrable conditionally independent random variables satisfyby the law of total covariance. Equal conditional means reduce this to the displayed variance; unequal conditional means need not give a nonnegative covariance. The equal-mean result explains correlated component evolutionary states in a coeval binary population without correlation between initial masses: variation in a shared environment can supply the entire covariance.

Condition on the common age $Y=y$ of the coeval binary population. Each component is a red giant with conditional probability $g(y)=y/100$, independently of the other component at that age. The conditional probability of at least one giant is $1-(1-g{)}^2=2g-g^2$. Integrating over the uniform age distribution givesThus the fraction is $299/30000=0.996\overline{6}\%$, just below $1\%$. The subtracted term removes the systems counted twice because both components are red giants. Using the square of the marginal giant fraction would incorrectly ignore their shared age.

A constant number star formation rate over $10\mathrm{Gyr}$ produces a uniform present-age distribution. The birth-age distribution under constant star formation therefore givesOutside this age interval the cumulative distribution function is zero or one. The census here retains stellar remnants as objects, as required by the population model.

Write $m=M/M_{\odot }$. The initial mass function is proportional to $m^{-3}$ on $m\ge 0.2$. Normalizing the tail integral gives the power-law initial-mass-function tail coordinate:For lower thresholds $X=1$. Because the birth mass and formation time have independence, the probability integral transform makes $(X,Y)$ uniform on the unit square. $X$ is an upper-tail coordinate, so larger masses have smaller $X$.

A red giant of current age $t$ lies in the stellar lifetime regions in a mass-age diagramThere are no red giants below $m=\sqrt{0.8}$ in this model. For $\sqrt{0.8}<m<1$ the upper boundary is the population's age limit, while for $m\ge 1$ both lifetime boundaries are present. A white dwarf lies above $t/\mathrm{Gyr}=10/m^2$.

In uniform probability coordinates the main-sequence boundary is $Y=20X$ and the end of the giant phase is $Y=25X$. The giant region is therefore $20X\le Y<25X$, clipped to the unit square: it is the triangle with vertices $(0,0),(1/25,1),(1/20,1)$. Its area, and hence the individual giant fraction, isThe white dwarf region $Y\ge 25X$ is the triangle with vertices $(0,0),(0,1),(1/25,1)$, givingThe remaining $97.5\%$ are on the main sequence. Boundaries have zero probability and do not affect these counts. Treating every post-giant object as a white dwarf is a stipulated toy-model assumption; real high-mass stars have other stellar remnants.

For the binary calculations, use a coeval binary population: components share their birth time even though their birth masses are independently drawn and their subsequent evolution is independent. Hence $(X_1,X_2,Y)$ is uniform in the unit cube. At a fixed age coordinate $Y=y$, a component is a giant when $y/25<X<y/20$, and a white dwarf when $0<X<y/25$. Define the conditional state fractionsComponents have conditional independence given $y$; averaging over the common age must be done after forming the conditional pair probabilities. The continuous-birth binary state fractions below are consequently volumes in this cube, not products of the marginal single-star fractions.

Let $T=10\mathrm{Gyr}$ and let $t$ denote present stellar age, rather than time measured from the galaxy's birth. With a fixed initial mass function, constant star formation rate also gives a constant number of births per unit time, as in the birth-age distribution under constant star formation. Counting the supplied stellar evolution model's remnants as part of the population, ages have a uniform distribution on $[0,T]$. Integrating the constant birth rate therefore givesOutside this interval the cumulative fraction is zero or one as appropriate.

Write $m=M/M_{\odot }$ for the birth-mass parameter and $M_{\min }=0.2M_{\odot }$. The normalization cancels when integrating the initial mass function:Thus the mass-tail fraction isIt is one at and below the lower cutoff. The normalization constant has no effect on any number fraction.

For evolutionary-state selection by stellar lifetimes, a red giant has completed its main sequence lifetime but not its giant lifetime. Its stellar lifetime regions in a mass-age diagram are bounded byThe lower boundary meets $t=T$ at $m=\sqrt{0.8}$ and the upper boundary meets it at $m=1$. A white dwarf lies above the upper lifetime curve, within the same age interval.

The probability integral transform, applied to the decreasing mass-tail coordinate, makes $X$ uniform on $(0,1)$: $\mathrm{Pr}(X<x)=x$. The independent birth age gives $Y=t/T$ uniform on $(0,1)$, so number fractions are areas in a unit square. Since $m^{-2}=25X$, the red giant region becomes $20X<Y<25X$, a triangle with vertices $(0,0)$, $(1/25,1)$ and $(1/20,1)$. Its area is $\frac{1}{2}(1/20-1/25)=1/200$. The white dwarf region $Y>25X$ is a triangle of area $\frac{1}{2}(1/25)=1/50$. Hence the present individual-star fractions areThe rest are on the main sequence under the given toy lifetime model.

For the binary stars, draw two independent mass-tail coordinates $X_1,X_2$, but only one shared age coordinate $Y$: components born together are coeval. The coeval binary population therefore occupies a uniform unit cube $(X_1,X_2,Y)$. At a fixed age $Y=y$, the state probabilities for either component areTheir evolutionary states have conditional independence given $y$. They generally do not have unconditional independence, because sharing an age correlates their states. The binary fractions below are slice areas averaged over $0<y<1$.

For the three unordered evolved pairs, conditional independence gives the slice probabilities $g^2$, $2gw=8g^2$ and $w^2=16g^2$. Averaging over the shared age in the coeval binary population yieldsThus their system-number ratio is $1:8:16$, as required.

For the unheaded instantaneous-burst continuation, put $\tau =t/\mathrm{Gyr}$, with $0<\tau <10$. All binary stars now have the same age, so the individual red giant fraction is $g=\tau /1000$. Independent component masses give the giant-pair fraction in an instantaneous stellar burst, the two-giant system fractionThe final comparison has an ambiguity about which population is being counted. Taken literally as a comparison with the first galaxy's individual-star giant fraction $0.005$, it would require $\tau >\sqrt{5000}\simeq 70.71$. There is no allowed burst age: throughout $0<\tau <10$, the two-giant fraction is less than $10^{-4}$, well below $0.005$.

If the intended comparison instead concerns two-giant systems in both galaxies, the comparison fraction is $1/30000$. Then ${\tau }^2/10^6>1/30000$ givesFor completeness, a comparison of individual red giant fractions in both galaxies would give $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. These three comparisons have different denominators and should not be conflated.

At a fixed age, the probability that at least one component is a red giant is $1-(1-g{)}^2=2g-g^2$. Integrating over the common age in the coeval binary population givesThis is just under one percent of systems. Squaring the marginal single-star fraction would incorrectly give two independently sampled component ages.

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.