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In what units are coefficients in survreg's dist=exponential?

I'm getting the following output:

Call:
survreg(formula = Surv(x, delta) ~ z, data = expdatauusi, dist = "exponential")
             Value Std. Error     z        p
(Intercept)  0.876      0.431  2.04 0.041835
z           -2.112      0.639 -3.30 0.000952

Scale fixed at 1 

Exponential distribution
Loglik(model)= -17.6   Loglik(intercept only)= -22.5
    Chisq= 9.65 on 1 degrees of freedom, p= 0.0019 
Number of Newton-Raphson Iterations: 4 
n= 30 
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In survival regression, the exponential and more generally, the Weibull distribution, is parametererized in its log-location-scale representation. When $T\sim\operatorname{Weibull}(\theta,\alpha)$ such that $$ S_T(t) = P(T>t) = e^{-(t/\theta)^\alpha}, $$ the log of $T$ belongs to the the location-scale family of distributions since \begin{align} S_{\ln T}(x)&=P(\ln T > x)\\ &=P(T>e^x) \\ &=e^{-(e^x/\theta)^\alpha} \\ &=e^{-e^\frac{x-\ln\theta}{1/\alpha}}. \end{align} From this we also see that $-\ln T$ is Gumbel distributed with location parameter $\mu=\ln\theta$ and scale parameter $\sigma=1/\alpha$. In the output from survreg the intercept plus z times the regression coefficients for z equals $\mu$ and the Scale parameter equals $\sigma$.

While $\theta$ has the same units as $T$ (say, seconds s), the parameter $\mu=\ln\theta$ is dimensionless although it is not very explicit that what we mean by $\ln\theta$ is really $\ln(\theta/1\text{s})$, such that we always get dimensionless quantities before taking logs. See related thread at math.stackexchange.

The scale parameter $\sigma$ is clearly also dimensionless (and equal to 1 in the exponential case).

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Edit to account for @Jarle Tufto's comment.

Survival regression models available through survreg are of location-scale form for some transformation of time (usually the log) and are not parametric proportional hazard models in general .

The exponential and Weibull distributions are exceptions.

Below I show how to transform the parameters from survreg(dist = 'exponential') in the proportional hazards setting to facilitate the interpretation.


In its most common form, the proportional hazards model is written $$ h(t) = h_0(t) \exp(\beta x) $$

Assuming an exponential distribution for the event times, $h_0(\cdot)$ is constant in time:

$$ h_0(t) = \lambda $$

To obtain estimates for the parameters $\lambda$ and $\beta$ from the R output, you can use the following transformations (only valid for exponential distribution): $$ \lambda = \exp\Big(-\text{(Intercept)}\Big) $$ $$ \beta = - z $$

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  • $\begingroup$ -1: Survival regression models available through survreg are of location-scale form for some transformation of time (usually the log) and are not parametric proportional hazard models in general (the Weibull and exponential model are more of exceptions that happen to have that property). $\endgroup$ – Jarle Tufto Mar 27 '17 at 10:59
  • $\begingroup$ @JarleTufto: There is no contradiction between my answer and your comment... Furthermore, the question is explicitly about the exponential distribution. And it is made explicit in my answer that it is only valid for the exponential distribution. However, I have edited my answer to account for your comment. $\endgroup$ – ocram Mar 27 '17 at 11:23

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