COV: *cov*ariance matrix for survival data

Description Usage Arguments Details Value Note See Also Examples

Description

covariance matrix for survival data

Usage

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COV(x, ...)

## S3 method for class 'ten'
COV(x, ..., reCalc = FALSE)

## S3 method for class 'stratTen'
COV(x, ..., reCalc = FALSE)

## S3 method for class 'numeric'
COV(x, ..., n, ncg)

Arguments

x

A numeric vector of number of events, e[t]. These are assumed to be ordered by discrete times.
A method is available for objects of class ten.

...

Additional arguments (not implemented).

reCalc

Recalcuate the values?
If reCalc=FALSE (the default) and the ten object already has the calculated values stored as an attribute, the value of the attribute is returned directly.

–Arguments for the numeric method:

n

number at risk (total).

ncg

number at risk, per covariate group.
If there are 2 groups, this can be given as a vector with the number at risk for group 1.
If there are >= 2 groups, it is a matrix with one column for each group.

Details

Gives variance-covariance matrix for comparing survival data for two or more groups.
Inputs are vectors corresponding to observations at a set of discrete time points for right censored data, except for n1, the no. at risk by predictor.
This should be specified as a vector for one group, otherwise as a matrix with each column corresponding to a group.

Value

An array.
The first two dimensions = the number of covariate groups K, k = 1, 2, … K. This is the square matrix below.
The third dimension is the number of observations (discrete time points).

To calculate this, we use x (= e[t] below) and n1, the number at risk in covariate group 1.
Where there are 2 groups, the resulting sparse square matrix (i.e. the non-diagonal elements are 0) at time t has diagonal elements:

cov[t] = - n0[t] * n1[t] * e[t] * (n[t] - e[t]) / (n[t]^2 * (n[t] - 1))

For >=2 groups, the resulting square matrix has diagonal elements given by:

cov[k, k, t] = n[k, t] * (n[t] - n[k, t]) * e[t] * (n[t] - e[t]) / (n[t]^2 * (n[t] - 1))

The off diagonal elements are:

cov[k, l, t] = - n[k, t] * n[l, t] * e[t] * (n[t] - e[t]) / n[t]^2 * (n[t] - 1)

Note

Where the is just one subject at risk n=1 at the final timepoint, the equations above may produce NaN due to division by zero. This is converted to 0 for simplicity.

See Also

Called by comp

The name of the function is capitalized to distinguish it from:
?stats::cov

Examples

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## Two covariate groups
## K&M. Example 7.2, pg 210, table 7.2 (last column).
data("kidney", package="KMsurv")
k1 <- with(kidney,
           ten(Surv(time=time, event=delta) ~ type))
COV(k1)[COV(k1) > 0]
## Four covariate groups
## K&M. Example 7.6, pg 217.
data("larynx", package="KMsurv")
l1 <- ten(Surv(time, delta) ~ stage, data=larynx)
rowSums(COV(l1), dims=2)
## example of numeric method
## Three covariate groups
## K&M. Example 7.4, pg 212.
data("bmt", package="KMsurv")
b1 <- asWide(ten(Surv(time=t2, event=d3) ~ group, data=bmt))
rowSums(b1[, COV(x=e, n=n, ncg=matrix(data=c(n_1, n_2, n_3), ncol=3))], dims=2)

dardisco/survMisc documentation built on May 14, 2019, 6:08 p.m.