fix(parametric-survival): use \vzero instead of \v0 so the PDF compiles

chapters/parametric-survival-models.qmd

@@ -355,14 +355,14 @@ Exponential regression models the hazard rate using a log link function:
 $$
 \ba
 \haz(t \mid \vx) &\eqdef \expf{\eta(\vx)} && \text{(exponential regression hazard specification)} \\
-\haz_0(t) &\eqdef \haz(t \mid \v0) && \text{(baseline hazard rate definition at $\vx = \v0$)} \\
-&= \expf{\eta(\v0)} && \text{(substituting $\vx = \v0$ into hazard function)} \\
+\haz_0(t) &\eqdef \haz(t \mid \vzero) && \text{(baseline hazard rate definition at $\vx = \vzero$)} \\
+&= \expf{\eta(\vzero)} && \text{(substituting $\vx = \vzero$ into hazard function)} \\
 &= \expf{\beta_0 + (\beta_1 \cdot 0 + \dots + \beta_p \cdot 0)} && \text{(evaluating linear predictor at zero vector)} \\
 &= \expf{\beta_0} && \text{(simplifying exponent)}
 \ea
 $$
 
-When there are no additional predictors ($\vx = \v0$), the baseline hazard is constant: $\lambda = \expf{\beta_0}$.
+When there are no additional predictors ($\vx = \vzero$), the baseline hazard is constant: $\lambda = \expf{\beta_0}$.
 This model uses a log link as in a generalized linear model for Poisson or rate outcomes.
 Because the hazard rate does not depend on time $t$, the hazard ratio comparing two covariate vectors $\vx$ and $\vxs$ is constant over time:
 
@@ -401,7 +401,7 @@ $$
 \surv(t \mid \vx) \eqdef \surv_0\left( t \cdot \hazfactor(\vx) \right)
 $$
 
-where $\hazfactor(\vx) \eqdef \expf{\eta(\vx)}$, $\eta(\vx) \eqdef \beta_1 x_1 + \dots + \beta_p x_p$, and $\surv_0(t) \eqdef \P(T \ge t \mid \vx = \v0)$ is the baseline survival function.
+where $\hazfactor(\vx) \eqdef \expf{\eta(\vx)}$, $\eta(\vx) \eqdef \beta_1 x_1 + \dots + \beta_p x_p$, and $\surv_0(t) \eqdef \P(T \ge t \mid \vx = \vzero)$ is the baseline survival function.
 
 We derive the expected survival time under the AFT model using change-of-variables integration:
 
@@ -411,7 +411,7 @@ $$
 &= \int_{0}^{\infty} \surv_0\left( t \cdot \hazfactor(\vx) \right) dt && \text{(substituting AFT survival function)} \\
 &= \int_{0}^{\infty} \surv_0(u) \cdot \frac{du}{\hazfactor(\vx)} && \text{(substituting $u = t \cdot \hazfactor(\vx) \implies dt = \frac{du}{\hazfactor(\vx)}$)} \\
 &= \hazfactor(\vx)^{-1} \cdot \int_{0}^{\infty} \surv_0(u) du && \text{(factoring constant $\hazfactor(\vx)^{-1}$ out of integral)} \\
-&= \hazfactor(\vx)^{-1} \cdot \E{T \mid \vx = \v0} && \text{(substituting baseline expectation formula)}
+&= \hazfactor(\vx)^{-1} \cdot \E{T \mid \vx = \vzero} && \text{(substituting baseline expectation formula)}
 \ea
 $$