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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 $0<y<1$.