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Here is one definition of the proportional odds model: $$\frac{S(t;{\bf Z})}{1-S(t;{\bf Z})}=e^{-{\bf Z}^T\pmb{\beta}}\frac{S_0(t;{\bf Z})}{1-S_0(t;{\bf Z})},$$ where ${\bf Z}$ = covariates, $\pmb{\beta}$ = regression coefficients, $T$ = survival time, $S(\cdot)$ = survival function, $S_0(\cdot)$ = baseline survival function. This equation is clearly equivalent to $\mbox{logit}[S(t;{\bf Z})]=\mbox{logit}[S_0(t)]-{\bf Z}^T\pmb{\beta}$.

Let $H(t)=-\mbox{logit}[S_0(t)]$. Now the book I'm reading claims, "If $H(t)$ is strictly increasing, [the first equation] can be equivalently written as $H(T)=-{\bf Z}^T\pmb{\beta}+W$, where the random variable $W$ follows a standard logistic distribution".

Can someone explain to me why this statement is true? First of all, I know that by definition $H(t)$ is monotone increasing, but I don't see why we need $H(t)$ to be strictly increasing. Then, it seems to me like the part where he says "$W$ follows a standard logistic distribution" is just plain wrong; why in the world should $-\mbox{logit}[S(t;{\bf Z})]$ follow a logistic distribution?!

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I think what he means is that you can rescale $t$ to make it a standard logistic random variable. First recall that if $F$ is the cdf of a random variable $X$, and $U$ is uniform, then $F^{-1}(U)$ is distributed like $X$ (where the inverse is disambiguated by using an infimum).

In other words, let $T:=S_0^{-1}(U)$, so that $\mbox{logit}(S_0(T))=\mbox{logit}(U)$, which is exactly $F^{-1}(U)$ where $F$ is the logistic cdf. Note that to be able to take the inverse, $S_0$ needs to be invertible, which is only possible when it's monotonic.The fact that $S_0$ is monotonic is implied from the assumption that $H(t)$ is monotonic, since the logit function is monotonic.

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  • $\begingroup$ I understand the first paragraph, but I don't completely understand the second. Can you please state, step-by-step, why $-\mbox{logit}[S(t;{\bf z})]$ follows a logistic distribution? $\endgroup$ – xFioraMstr18 Jun 22 '17 at 2:41
  • $\begingroup$ en.m.wikipedia.org/wiki/Logistic_distribution $\endgroup$ – Alex R. Jun 22 '17 at 3:14
  • $\begingroup$ I still don't understand. $\endgroup$ – xFioraMstr18 Jun 22 '17 at 17:33

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