docs(proportional-hazards): improve narrative flow and fix line breaks

Refine text explanations, resolve spatial language ("below" -> "at or prior to"), and enforce semantic line breaks across Proportional Hazards subfiles.

Co-authored-by: dem-extra1 <112029334+dem-extra1@users.noreply.github.com>

Author: google-labs-jules[bot]

_subfiles/proportional-hazards-models/_lem-ph-breslow-point-mass.qmd

@@ -32,7 +32,7 @@ But the hazard-density factor $\blue{\haz_0(t_i)}^{\delta_j}$ in @eq-ph-breslow-
 is maximized when all of that mass is concentrated as a single point at $t_i$
 (so $\blue{\haz_0(t_i)} = \blue{h_{0i}}$):
 spreading the same total mass across a wider neighborhood would leave less of it exactly at $t_i$,
-lowering $\blue{\haz_0(t_i)}$ below $\blue{h_{0i}}$, while leaving the survival penalty fixed.
+lowering $\blue{\haz_0(t_i)}$ to less than $\blue{h_{0i}}$, while leaving the survival penalty fixed.
 Since concentrating the mass strictly increases the hazard-density factor
 without changing the survival penalty, @eq-ph-breslow-full-lik is maximized by @eq-ph-breslow-point-mass.
 :::
@@ -48,7 +48,7 @@ Splitting the same total mass into two halves —
 $0.05$ exactly at $t_1$ and $0.05$ at a nearby non-event point —
 leaves only $\blue{\haz_0(t_1)} = 0.05$ at $t_1$ itself, *half* the hazard-density factor,
 while the cumulative hazard $\blue{\cuhaz_0(\tilde T_j)}$ (and hence every survival term)
-is unchanged, since both allocations place the same total mass $0.1$ below any $\tilde T_j \ge t_1$.
+is unchanged, since both allocations place the same total mass $0.1$ at or prior to any $\tilde T_j \ge t_1$.
 Splitting the mass therefore strictly decreases the likelihood,
 confirming that concentrating all of it at $t_1$ is optimal.
 :::

_subfiles/proportional-hazards-models/_proof-breslow-baseline-cuhaz-est.qmd

@@ -9,7 +9,8 @@ Assume, as in the partial-likelihood proof, that there are no tied event times,
 so each ordered event time $t_i$ corresponds to exactly one event,
 and let $K$ denote the number of distinct event times $t_1 < \cdots < t_K$.
 
-Fix $\vb$. By @lem-ph-breslow-point-mass, the maximizing $\haz_0(\cdot)$
+Fix $\vb$.
+By @lem-ph-breslow-point-mass, the maximizing $\haz_0(\cdot)$
 places point masses $\blue{h_{0i}} \eqdef \blue{\haz_0(t_i)}$ only at the $K$ event times.
 By @lem-ph-breslow-profile-lik, substituting this form into @eq-ph-breslow-full-lik
 gives the profile likelihood @eq-ph-breslow-profile-lik in terms of the point masses

_subfiles/proportional-hazards-models/_sec-exm-coxph-bmt.qmd

@@ -13,7 +13,8 @@ The table provides hypothesis tests comparing groups 2 and 3 to group 1.
 Group 3 has the highest hazard, so the most significant comparison is
 not directly shown.
 
-The coefficient `r round(coef(bmt.cox)["groupHigh Risk AML"], 4)` is on the log-hazard-ratio scale. The next column gives the hazard ratio `r round(exp(coef(bmt.cox)["groupHigh Risk AML"]), 4)`, and a hypothesis (Wald) test.
+The coefficient `r round(coef(bmt.cox)["groupHigh Risk AML"], 4)` is on the log-hazard-ratio scale.
+The next column gives the hazard ratio `r round(exp(coef(bmt.cox)["groupHigh Risk AML"]), 4)`, and a hypothesis (Wald) test.
 
 The (not shown) group 3 vs. group 2 log hazard ratio is
 `r round(coef(bmt.cox)["groupHigh Risk AML"], 4)` − (`r round(coef(bmt.cox)["groupLow Risk AML"], 4)`) =
@@ -79,7 +80,9 @@ list(KM = km_fit, Cox = cox_fit) |>
 Survival Functions for Three Groups by KM and Cox Model
 :::
 
-When we use `survfit()` with a Cox model, we have to specify the covariate levels we are interested in; the argument `newdata` should include a `data.frame` with the same named columns as the predictors in the Cox model and one or more levels of each.
+When we use `survfit()` with a Cox model,
+we have to specify the covariate levels we are interested in;
+the argument `newdata` should include a `data.frame` with the same named columns as the predictors in the Cox model and one or more levels of each.
 
 {{< slidebreak >}}
 

_subfiles/proportional-hazards-models/_sec-test-ph-assumption.qmd

@@ -2,8 +2,10 @@
 ::: notes
 The Nelson-Aalen estimate of the cumulative hazard is usually used for estimates of the hazard and often the cumulative hazard.
 
-If the hazards of the three groups are proportional, that means that the ratio of the hazards is constant over $t$. We can test this using the ratios of the estimated cumulative hazards, which also would be
-proportional.
+If the hazards of the three groups are proportional,
+that means that the ratio of the hazards is constant over $t$.
+We can test this using the ratios of the estimated cumulative hazards,
+which also would be proportional.
 :::
 
 ```{r}
@@ -82,7 +84,7 @@ Hazard Ratios by Disease Group (0-300 Days)
 
 {{< slidebreak >}}
 
-The cumulative hazard curves should also be proportional
+The cumulative hazard curves should also be proportional.
 
 :::{#fig-cuhaz-bmt}
 ```{r}
@@ -119,13 +121,15 @@ Disease-Free Cumulative Hazard by Disease Group (log-scale)
 
 ::: notes
 The Nelson-Aalen estimate of the cumulative hazard is usually used for
-estimates of the hazard. Since the hazard is the derivative of the
-cumulative hazard, we need a smooth estimate of the cumulative hazard,
+estimates of the hazard.
+Since the hazard is the derivative of the cumulative hazard,
+we need a smooth estimate of the cumulative hazard,
 which is provided by smoothing the step-function cumulative hazard.
 
-The R package `muhaz` handles this for us. What we are looking for is
-whether the hazard function is more or less the same shape, increasing,
-decreasing, constant, etc. Are the hazards "proportional"?
+The R package `muhaz` handles this task.
+What we are looking for is whether the hazard function is more or less the same shape,
+increasing, decreasing, constant, etc.
+Are the hazards "proportional"?
 :::
 
 

_subfiles/proportional-hazards-models/_sec-understand-coxph.qmd

@@ -207,7 +207,9 @@ it is called the **proportional hazards assumption**.
 ### proportional hazards
 
 A conditional probability distribution $p(T|X)$
-has **proportional hazards** if the hazard ratio $\haz(t|\vx_1)/\haz(t|\vx_2)$ does not depend on $t$. Mathematically, it can be written as:
+has **proportional hazards** if the hazard ratio $\haz(t|\vx_1)/\haz(t|\vx_2)$
+does not depend on $t$.
+Mathematically, it can be written as:
 
 $$
 \frac{\haz(t|\vx_1)}{\haz(t|\vx_2)}

_subfiles/proportional-hazards-models/_sec_coxph-adjust_ties.qmd

@@ -5,7 +5,8 @@ At each time $t_i$ at which more than one of the subjects has an event,
 let $d_i$ be the number of events at that time, $D_i$ the set of
 subjects with events at that time, and let $s_i$ be a covariate vector
 for an artificial subject obtained by adding up the covariate values for
-the subjects with an event at time $t_i$. Let
+the subjects with an event at time $t_i$.
+Let
 $$\bar\eta_i = \beta_1s_{i1}+\cdots+\beta_ps_{ip}$$ and
 $\bar\theta_i = \expf{\bar\eta_i}$.
 
@@ -52,10 +53,11 @@ regression.
 #### Example: Breslow's method
 
 Suppose as an example we have a time $t$ where there are 20 individuals
-at risk and three failures. Let the three individuals have risk
+at risk and three failures.
+Let the three individuals have risk
 parameters $\theta_1, \theta_2, \theta_3$ and let the sum of the risk
-parameters of the remaining 17 individuals be $\theta_R$. Then the
-factor in the partial likelihood at time $t$ using Breslow's method is
+parameters of the remaining 17 individuals be $\theta_R$.
+Then the factor in the partial likelihood at time $t$ using Breslow's method is
 
 ::: smaller
 $$
@@ -65,7 +67,7 @@ $$
 $$
 :::
 
-If on the other hand, they had died in the order 1,2, 3, then the
+If on the other hand, they had died in the order 1, 2, 3, then the
 contribution to the partial likelihood would be:
 
 ::: smaller
@@ -76,15 +78,14 @@ $$
 $$
 :::
 
-as the risk set got smaller with each failure. The exact method roughly
-averages the results for the six possible orderings of the failures.
+as the risk set got smaller with each failure.
+The exact method roughly averages the results for the six possible orderings of the failures.
 
 #### Example: Efron's method
 
-But we don't know the order they failed in, so instead of reducing the
-denominator by one risk coefficient each time, we reduce it by the same
-fraction.
-This fractional reduction approach is Efron's method.
+Because the exact failure order among tied event times is unobserved,
+rather than reducing the denominator by one risk coefficient at each step,
+Efron's method reduces it by a uniform fractional amount.
 
 ::: smaller
 $$\left(\frac{\theta_1}{\theta_R+\theta_1+\theta_2+\theta_3}\right)