Prediction in Cox Proportional-Hazard Regression

He,
I used the "coxph" function, with four covariates.

Let's say something like that

model.1 <- coxph(Surv(Time,Event)~X1+X2+X3+X4,data=DATA)

So I obtain the 4 coefficients B1,B2,B3,B4 such that

h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4).

When I use the function on the same data

predict.coxph(model.1,type="lp")

how it works in making the prediction?
I mean which is the formula, given the data-point P1=[X1(1),X2(1),X3(1),X4(1)],
that the function "predict.coxph" use to make the prediction of P1.

From the code (getAnywhere("predict.coxph"))

On Thu, 9 Jun 2005 Giuseppe.Palermo at bo.infn.it wrote:

He,
I used the "coxph" function, with four covariates.

Let's say something like that

model.1 <- coxph(Surv(Time,Event)~X1+X2+X3+X4,data=DATA)

So I obtain the 4 coefficients B1,B2,B3,B4 such that

h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4).

When I use the function on the same data

predict.coxph(model.1,type="lp")

How does that work?  predict.coxph is not an exported function!

how it works in making the prediction?
I mean which is the formula, given the data-point 
P1=[X1(1),X2(1),X3(1),X4(1)],
that the function "predict.coxph" use to make the prediction of P1.

From the code (getAnywhere("predict.coxph"))

    if (type == "lp" || type == "risk") {
        if (missing(newdata)) {
            pred <- object$linear.predictors
            names(pred) <- names(object$residuals)
        }
        else pred <- x %*% coef + offset
...

so that is the formula it uses.  As you did not supply 'newdata', it 
quotes the 'linear.predictors' component of the fit: see 
?coxph.object.

Effectively it centred the explanatory variables on their means and 
then applied the linear regression formula to give the linear 
predictor. It is the centring that may be non-obvious: effectively 
h_0(t), the baseline hazard, is taken at the average of the subjects.

Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

Quoting Prof Brian Ripley <ripley at stats.ox.ac.uk>:

On Thu, 9 Jun 2005 Giuseppe.Palermo at bo.infn.it wrote:

He,
I used the "coxph" function, with four covariates.

Let's say something like that

model.1 <- coxph(Surv(Time,Event)~X1+X2+X3+X4,data=DATA)

So I obtain the 4 coefficients B1,B2,B3,B4 such that

h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4).

When I use the function on the same data

predict.coxph(model.1,type="lp")

How does that work?  predict.coxph is not an exported function!

how it works in making the prediction?
I mean which is the formula, given the data-point 
P1=[X1(1),X2(1),X3(1),X4(1)],
that the function "predict.coxph" use to make the prediction of P1.

From the code (getAnywhere("predict.coxph"))

    if (type == "lp" || type == "risk") {
        if (missing(newdata)) {
            pred <- object$linear.predictors
            names(pred) <- names(object$residuals)
        }
        else pred <- x %*% coef + offset
...

so that is the formula it uses.  As you did not supply 'newdata', it quotes 
the 'linear.predictors' component of the fit: see ?coxph.object.

Effectively it centred the explanatory variables on their means and then 
applied the linear regression formula to give the linear predictor. It is 
the centring that may be non-obvious: effectively h_0(t), the baseline 
hazard, is taken at the average of the subjects.

Dear Prof. Ripley
Thanks for replying to me email.
I only have an other question:

since h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4)
represent the hazard at time t.

In a linear prediction,
what     Value = B1*(X1-mean(X1)) + B2*(X2-mean(X2)) + ....
represent?

I only have an other question:

since h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4)
represent the hazard at time t.

In a linear prediction,
what     Value = B1*(X1-mean(X1)) + B2*(X2-mean(X2)) + ....
represent?