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

Question

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 

Answer 1

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).

Answer 2

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.

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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): $$ \lambda = \exp\Big(-\text{(Intercept)}\Big) $$ $$ \beta = - z $$