Thanks for your advise, David.
I did read the help for survreg and using the followings to calculate.

survreg's scale  =    1/(rweibull shape)

survreg's intercept = log(rweibull scale)

However, the scale in rweibull has been transformed by exp(betaX's). In my
case, the baseline hazard for T is 0.004. I'm wondering if it's possible to
get the estimated baseline hazard by the output from survreg.
Thank you very much.

On Tue, Mar 6, 2012 at 5:06 PM, David Winsemius <dwinsemius@comcast.net>wrote:

>
> On Mar 6, 2012, at 5:53 PM, FU-WEN LIANG wrote:
>
>  Hi all,
>>
>> I'm trying to generate a Weibull distribution including four covariates in
>> the model. Here is the code I used:
>>
>> T = rweibull(200, shape=1.3,
>> scale=0.004*exp(-(-2.5*b1+2.5***b2+0.9*x1-1.3*x2)/1.3))
>> C = rweibull(n, shape=1.5, scale=0.008)   #censoring time
>> time = pmin(T,C)  #observed time is min of censored and true
>> event = time==T   # set to 1 if event is observed
>> return(cbind(time,event,T,C))
>>
>> And then, I used this dataset to fit a Weibull regression mode:
>>
>> survreg(Surv(time, event)~b1+b2+x1+x2, dist='weibull')
>>
>> The result I got was:
>> Intercept       b1         b2         x1          x2       Scale
>> -3.983       1.916    -1.828    -0.808    1.022       0.778
>>
>> I transformed those estimates into:
>>
>> Shape = 1/0.778=1.3
>> beta1 = -1.916/0.778 = -2.5
>> beta2 = -(-1.828)/0.778 = 2.5
>> beta3 = -(-0.808)/0.778 = 1.0
>> beta4 = -1.022/0.778 = -1.3
>> scale =  exp(-3.983) = 0.0186
>>
>>  From the result, we can see all parameter estimates are close to the
>>>
>> initial values, except for scale. The baseline scale is 0.004.
>> The scale parameter is varying by covariates.
>> If so, how can I double check if the estimated baseline scale is close to
>> the initial baseline scale?
>>
>
> Please read the read the help() for 'survreg' and make sure you follow the
> links provided and work through the examples.
>
>
>
>  Please advise if my concept
>>
>
> The concept is the problem.
>
>
>
>  or code is not correct.
>>
>> Thank you very much.
>>
>
> --
>
> David Winsemius, MD
> West Hartford, CT
>
>

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