Kernel-smoothed baseline hazard

\(\lambda_{n0}(y) = h_n^{-1}\int K((y - \xi)/h_n) d\Lambda_{n0}(\xi)\) (6.44), with \(\Lambda_{n0}\) the Breslow estimator of the cumulative baseline hazard.

Usage

morie_baseline_hazard(t, x, event, beta_hat, bandwidth = NULL, grid = NULL)

Arguments

t

observed durations, non-negative.

x

numeric matrix of covariates.

event

1 for an event, 0 for right-censoring.

beta_hat

regression coefficients.

bandwidth

h_n; a spread-scaled n^(-1/5) when NULL.

grid

evaluation points.

Value

list: grid, lambda0_hat, cumhaz_times, cumhaz, bandwidth, A_K, rate_exponent, root_n_attainable, n_events, n, method.

Details

The smoothing is not cosmetic. \(\lambda_0 = d\Lambda_0/dy\), so the obvious estimator is \(d\Lambda_{n0}/dy\) – and that does not work, because \(\Lambda_{n0}\) is a STEP function whose derivative is zero almost everywhere and undefined at the jumps. Same obstruction as differentiating an empirical distribution function to get a density, same remedy: smooth first.

The rate is no faster than \(n^{-2/5}\) for a twice differentiable \(\lambda_0\) and a second-order kernel; \(n^{-1/2}\) is NOT attainable. The leading bias is \(A_K h_n^2 \lambda_0''(y)/2\) with \(A_K = \int \zeta^2 K(\zeta)d\zeta\), returned so the bias is computable. Mirrors morie.fn.hrzlam.

References

Horowitz, Sec. 6.2.4, eqs. (6.44)-(6.48).

Examples

set.seed(1)
n <- 200
x <- cbind(rnorm(n), rnorm(n))
tt <- rexp(n) / exp(x %*% c(0.5, -0.3))
morie_baseline_hazard(tt, x, rep(1, n), c(0.5, -0.3))$root_n_attainable
#> [1] FALSE