If I run a Weibull survival model in R with the code
survreg(Surv(t,delta)~expalatory variables, dist="w")
how do I interpret the output of the model? That is, is the form of the model just $1-\exp(e/\lambda)^k)$ with $\lambda$ the scale and $k$ the shape or does it take a different form?
I have found something which says that the output is of the form $$\exp(-\exp(-\alpha_0-\alpha_1-\ldots)^kx^k),$$ where the $\alpha_i$ are the coefficients of the covariates. If so, the output would give me the following parameters:
$$k=k \quad \mbox{and} \quad \lambda=\frac{1}{\exp(-\alpha_0-\alpha_1\ldots)}.$$
Best Answer
Ok so I'm just going to post an answer here using the R help that DWin described. Using the function
rweibullin R gives the usual form of the Weibull distribution, with its cumulative function being:$$F(x)=1-\exp(-\left ( \frac{x}{b}\right )^a)$$
So we will denote the shape parameter of
rweibullby $a$ and the scale parameter ofrweibullby $b$.Now the problem is that the output of
survreggives both shape and scale parameters which are not the same as the shape and scale parameters fromrweibull. Let us denote the shape parameter fromsurvregas $a_s$ and the scale parameter ofsurvregby $b_s$.Then, from
?survregwe have that:So this gives us that:
$$a=\frac{1}{b_s}\quad \mbox{and} \quad b=\exp(a_s)$$
So if we suppose that we run the function
survregwith $n$ covariates, then the output will be:$\alpha_0,\ldots, \alpha_{n-1}$, the coefficents of the covariates and some scale parameter $k$. The Weibull model given in standard form is then given by:
$$F(x)=1-\exp\left (- \left (\frac{x}{\exp(\alpha_0+\alpha_1+\ldots +\alpha_{n-1})} \right ) ^{\frac{1}{k}}\right )$$