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.

The additive Abelian group $G$ is divisible because it is a real vector space. Each convex subgroup $C_n$ is divisible too: for $c\ge 0$ in $C_n$ and a positive integer $m$, its unique additive quotient $c/m$ satisfies $0\le c/m\le c$, hence belongs to $C_n$. Negative elements follow by taking inverses. Thus all $C_n$ are rational vector subspaces, independently of whether the given real scalar multiplication respects the order.

Choose an ordered-group isomorphism ${\theta }_n:C_{n+1}/C_n\to \mathbb{R}$. It is rational-linear, since additive homomorphisms between divisible torsion-free groups preserve division by integers. Extend a rational basis of $C_n$ to one of $C_{n+1}$; equivalently choose a rational-linear section $s_n:\mathbb{R}\to C_{n+1}$ of ${\theta }_n$ composed with the quotient map. ThenSince $C_1=0$ and the union is $G$, each $g\in G$ has a unique expression $g={\sum }_{j=1}^N{s}_j(r_j)$ with only finitely many nonzero coefficients. Define $\Phi (g)=(r_1,r_2,\dots )$. This is an injective additive group homomorphism. If $r_N\ne 0$ is the largest active coefficient, the image of $g$ in the ordered quotient $C_{N+1}/C_N$ has sign equal to that of $r_N$. Convexity ensures that this is also the sign of $g$.

The maximum-support convention for the Hahn group $V({\mathbb{Z}}_{>0},\mathbb{R})$ uses precisely finite supports and the sign of the largest active coefficient. Hence $\Phi$ preserves and reflects the total order, and therefore preserves join and meet. This proves the finite-support Hahn embedding along a countable convex chain. In fact every finite coefficient sequence is obtained, so $\Phi$ is an ordered-group isomorphism in this convention; the requested embedding follows.

The Ohnishi orderability criterion states that, for each finite list $g_1,\dots ,g_n\ne 1$, there are signs ${\epsilon }_i\in \{-1,1\}$ such that the semigroup of nonempty products of conjugates of $g_i^{{\epsilon }_i}$ does not contain $1$. The conjugates are essential for a two-sided invariant total order. Necessity follows by choosing all signed elements positive: their conjugates and nonempty products are still positive.

For sufficiency, consider the compact product $\{+,-{\}}^{G\setminus \{1\}}$. Impose the closed conditions that an inverse has the opposite sign, that conjugation preserves sign, and that a product of two positive elements is positive whenever it is nonidentity. A positive pair cannot multiply to the identity because of the inverse condition. Each condition uses finitely many coordinates. Given finitely many conditions, collect every nonidentity element appearing in them into a finite list and choose signs using the criterion. These signs satisfy the conditions: an incorrectly negative product would contribute its positive inverse to the generated semigroup, producing identity with the positive factors; an incorrect conjugate or inverse gives the same contradiction. Extend the signs arbitrarily to the remaining coordinates. Thus the finite intersection property and compactness give a global sign assignment. Its positive elements form a conjugation-invariant semigroup $S$ with $S\cap S^{-1}=\varnothing$ and $S\cup S^{-1}=G\setminus \{1\}$. Defining $a<b$ by $a^{-1}b\in S$ gives the required linearly ordered group.

Restriction shows that every finitely generated subgroup of an orderable group is orderable. Conversely, suppose every such subgroup is orderable. If the criterion failed on $g_1,\dots ,g_n$, each of the finitely many sign choices would have an identity witness made of finitely many conjugates. Collect all $g_i$ and all conjugating elements in these witnesses into one finitely generated subgroup $H$. An order on $H$ supplies one positive sign choice, but its corresponding witness is a nonempty product of positive elements in $H$, a contradiction. Hence the criterion holds in $G$.

Finally, a finitely generated subgroup of a torsion-free abelian group is ${\mathbb{Z}}^r$ by the Fundamental theorem of finitely generated abelian groups. Its lexicographic order is two-sided invariant. Applying the local-to-global conclusion proves that every torsion-free abelian group admits the required order.

The response is the species categorical variable, while the four measurements are candidate predictors. The dot in the formula means all other columns of the data frame. The fitted object is a classification tree: it recursively partitions predictor space by binary threshold rules and assigns a species and class-probability vector to each terminal region. Sepal width was available but was not selected in the displayed tree.

At a node $t$, let $n_{tk}$ be the number of observations of class $k$ and $n_t={\sum }_k{n}_{tk}$. The multinomial likelihood is maximized by ${\hat{p}}_{tk}=n_{tk}/n_t$. The fitted class is one attaining the largest proportion, with a software convention resolving ties. The classification-tree deviance isA pure node has zero deviance. At each nonterminal node, consider candidate split points between distinct observed predictor values, and choose the split with the largest decrease $D(t)-D(t_L)-D(t_R)$. Repeating this local maximization constructs a binary decision tree; it does not solve a global optimization over all trees. Growth stops when nodes are pure or size/deviance controls prohibit further splitting. Mixed leaves with five or six observations can therefore remain. The node number $j$ has children $2j$ and $2j+1$, and a terminal-node marker means that no further split was retained.

The probability triples are in class order $(c,s,v)$, not $(s,c,v)$. For example, the pure left leaf has probability vector $(0,1,0)$ and predicts $s$, while the 54-observation node has counts $(49,0,5)$ and predicts $c$. The root counts are $(50,50,50)$, giving $D=300\log 3\simeq 329.584$ and a tied majority class. Its reported choice of $c$ does not indicate a more common class.

The complete derived decision rule is shown below. Each leaf includes its fitted class and the observed class counts, so both the branching logic and its remaining errors are visible. For the observed data, the printed strict inequalities leave no equality cases because the thresholds fall between measurement values. A consistent right-branch convention at equality extends the rule to new specimens.

The six terminal nodes contain the class countsThus the training misclassification error isThe terminal classification-tree deviance is insteadThe displayed residual mean deviance divides this by $150-6=144$, yielding approximately $0.1253$. This is the summary's reporting convention, not a normal-error residual mean square or an automatic chi-squared calibration with 144 statistical degrees of freedom. The two summary measures answer different questions: deviance depends on the fitted probability of each observed class, while misclassification counts only whether the majority-class prediction is correct. Both are training quantities. Cross-validation or independent test data, rather than the displayed training error, are needed to assess prediction on new specimens; cost-complexity tree pruning can select a less elaborate tree if predictive performance warrants it.

For the numerical competing risks update, the estimated probability of remaining event-free just before the tied time isThere were no intervening events, so the cumulative incidence functions did not jump between the two stated times. The Aalen–Johansen estimator uses the risk set immediately before the event time, including the subject censored at that same time under the usual events-before-censoring tie convention. Therefore the cause-$A$ increment is $0.474(2/20)$, andFor a consistency check, ${\hat{F}}_B(7.8)=0.241+0.474/20=0.2647$ and $\hat{S}(7.8)=0.474(1-3/20)=0.4029$; these sum to one. The censoring does not create an incidence jump and leaves $16$ subjects for subsequent follow-up. Dividing by $19$ would instead assume that the censored subject had already left before the event time, contrary to the stated immediately preceding risk set. This is a cumulative incidence update with tied censoring.

Here is a finite-horizon Girsanov theorem statement with the sign fixed explicitly. Let $B$ be a Brownian motion under $P$, and let $\theta$ be predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Ifis a true martingale with $\mathbb{E}{L}_T=1$, then the measure $Q$ defined by $dQ=L_{T}dP$ makes $B_t-{\int }_0^t{\theta }_{s}ds$ a Brownian motion up to $T$. The sufficient Novikov condition is $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$. Merely being a local martingale is not a sufficient density condition.

For a deterministic absolutely continuous function $h$ with $h(0)=0$ and derivative $g\in L^2[0,T]$, take $\theta =g$. Novikov's condition holds because its exponential is deterministic and finite. Under $Q$, the canonical process has the law of a Brownian motion plus $h$. Thus for every bounded measurable path functional $F$,This derives the Cameron-Martin theorem density formula from the Girsanov theorem, rather than assuming the two measures have the same drift convention. Replacing $h$ by $-h$ changes the sign of the stochastic integral.

The question's path space is the whole half-line, not just one finite horizon. When $g\in L^2(0,\infty )$, the density martingale satisfiesIt has a uniformly integrable limit $L_{\infty }=\exp ({\int }_0^{\infty }{g}_{s}d{B}_s-\Vert g{\Vert }_2^2/2)$ of mean one, strictly positive almost surely. Integrals here converge in $L^2$ and almost surely. The finite-time identities then identify $L_{\infty }$ as the density on the full path sigma-algebra, since finite-time cylinder events generate it. This is Cameron-Martin shifts on infinite Wiener path space. Finite-horizon absolute continuity of measures alone does not imply infinite-horizon absolute continuity of measures.

Use $T$ for the fixed maturity and $s\le T$ for the observation time. The original PDF's discounting uses the instantaneous forward rate over $[s,T]$ and the short rate over $[0,s]$. Let $X(s,u)=F_{s,u}-{\mu }_{s,u}$ be the centered Gaussian forward-rate field, and defineAssume the covariance and mean have the regularity needed for these mean-square integrals and the maturity differentiation below. Splitting $Y$ at $u=s$ shows that the discounted zero-coupon bond price isThis integrated Gaussian forward-rate process has deterministic varianceIts independent increments hold relative to the entire specified filtration. Indeed, if $r\le s$ and $z\le r$, the covariance between $Y(s,T)-Y(r,T)$ and any earlier $X(z,w)$ is the integral ofSince uncorrelated jointly Gaussian variables are independent, the increment is independent of every finite family of earlier observations and hence of their generated sigma-algebra. The increment has normal distribution with mean zero and variance $v(s,T)-v(r,T)$. Also $Y(0,T)=0$ because $c(0,u,w)=0$.

The Gaussian moment-generating function now givesTherefore the Gaussian forward-rate covariance drift restriction is exactly the deterministic conditionIt turns $Z_{s,T}/Z_{0,T}$ into a Gaussian exponential martingale with deterministic variance. The subparts establish the requested equivalent descriptions from this common calculation.

Standard one-dimensional Brownian motion is a real process with $B_0=0$, continuous paths almost surely, and increments $B_t-B_s\sim N(0,t-s)$ independent of the past at time $s$, for $0\le s<t$. These conditions determine its entire finite-dimensional distribution family.

The Wiener theorem asserts that such a process exists, equivalently that there is a unique Wiener measure on $C_0([0,\infty ),\mathbb{R})$ under which the coordinate process has these properties. To construct it, prescribe centered multivariate normal laws with covariance $\mathbb{E}{B}_s{B}_t=\min (s,t)$. This covariance is positive semidefinite becauseThe laws are consistent under removing coordinates, so the Kolmogorov extension theorem constructs the process. Normal increments have fourth moment $\mathbb{E}|B_t-B_s{|}^4=3|t-s{|}^2$. The Kolmogorov continuity theorem therefore supplies a continuous modification, retaining all the prescribed finite-dimensional laws and hence the independent normal increments. Apply this on every compact time interval, taking consistent versions; continuity and equality at rational times make them agree on overlaps. The resulting law on continuous-path space is unique because evaluations at rational times generate its Borel sigma-algebra. This is a proof sketch of the existence theorem, independent of the random-walk limit in Question 5.