Calculating the Log Hazard Ratio with SE in R

Question

In R, is there a predefined function that will give me the log hazard ratio and its standard error for a black male (as shown in the example below) given the output of coxph regression?

library(survival)
library(KMsurv)

#Kidney transplant data from Klein and Moeshberger. Massage data to make
#results look like those in book
data(kidtran)
data2 <- kidtran
data2$Gender <- "male"
    data2[data2$gender==2,7] <- "female"
data2$Race <- "white"
    data2[data2$race==2,8] <- "black"
data2$Gender <- as.factor(data2$Gender)
data2$Race <- as.factor(data2$Race)
data2$Race <- relevel(data2$Race,ref="white")

fit2 <- coxph(Surv(time,delta) ~ Gender * Race, data=data2)
summary(fit2)

#Relative log risk for a black male (reference white female) from
#page 252 in Klein and Moeshberger
(coef(fit2)[3] + coef(fit2)[2] + coef(fit2)[1])
sqrt(sum(diag(fit2$var)) + 2*fit2$var[2,1] + 2*fit2$var[3,1] + 2*fit2$var[3,2])

#Let's use predict
black.male <- data.frame(
  Gender="male",
  Race="black"
)

white.female <- data.frame(
  Gender="female",
  Race="white"
)

bm <- predict(fit2,newdata=black.male,se.fit=TRUE)

#bm in terms of original coefficients
coef(fit2)[3]*(1-fit2$means[3]) + coef(fit2)[2]*(1-fit2$means[2]) + coef(fit2)[1]*(1-fit2$means[1])

wf <- predict(fit2,newdata=white.female,se.fit=TRUE)

#Relative log risk and se for a black male (reference white female) 
bm$fit - wf$fit
sqrt(bm$se.fit^2 + wf$se.fit^2)

Answer 1

I do not think there exists a function like logHazardRatio("black males", "white females") which would return an estimate and a standard error. However, you can easily recover them as below.

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Answer prior to your edit

Your Cox model is as follows, $$h_j(t) = h_0(t) \exp(\beta_1 \textrm{gender}_j + \beta_2 \textrm{race}_j + \beta_3 \textrm{gender}_j * \textrm{race}_j),$$ with $$\textrm{gender} = \left\{ \begin{array}{l} 0 \textrm{ if female} \\ 1 \textrm{ if male,} \\ \end{array} \right. $$ $$ \textrm{race} = \left\{ \begin{array}{l} 0 \textrm{ if white} \\ 1 \textrm{ if black,} \\ \end{array} \right.$$ and we have $$\frac{\textrm{hazard in black males}}{\textrm{hazard in white females}} = \frac{h_0(t) \exp(\beta_1 + \beta_2 + \beta_3)}{h_0(t)}$$ so that $$\log \left( \frac{\textrm{hazard in black males}}{\textrm{hazard in white females}} \right) = \beta_1 + \beta_2 + \beta_3.$$

Your estimate coef(fit2)[3] + coef(fit2)[2] + coef(fit2)[1] is thus correct :-)

To get a standard error, note that $$ \begin{align*} \textrm{Var}(\hat{\beta}_1 + \hat{\beta}_2 + \hat{\beta}_3) & = \textrm{Var}(\hat{\beta}_1) + \textrm{Var}(\hat{\beta}_2) + \textrm{Var}(\hat{\beta}_3) \\ & + 2 \textrm{Cov}(\hat{\beta}_1, \hat{\beta}_2) + 2 \textrm{Cov}(\hat{\beta}_1, \hat{\beta}_3) + 2 \textrm{Cov}(\hat{\beta}_2, \hat{\beta}_3). \end{align*} $$ Estimates for all of these terms can be found in the estimated covariance matrix of $(\hat{\beta}_1, \hat{\beta}_2, \hat{\beta}_3)'$:

> fit2$var
            [,1]        [,2]        [,3]
[1,]  0.03941978  0.02576211 -0.03945987
[2,]  0.02576211  0.09731841 -0.09729282
[3,] -0.03945987 -0.09729282  0.18242564

For example, $\widehat{\textrm{Cov}}(\hat{\beta}_1, \hat{\beta}_3) = -0.03945987.$