This is an attempt to prove the hazard function $h$ of the accelerated failure time model as presented by Klein and Moeschberger in "Survival Analysis: Techniques for Censored and Truncated Data," (2nd Edition, Springer, 2003) on page 47, Eqn 2.6.2:
$h(x|z) = h_0(x \exp(-\gamma^t z)) \exp(-\gamma^t z)$.
Here, $x$ is time, $z$ the vector of covariate values, $\gamma$ the corresponding coefficients, and $h_0$ the baseline hazard function when all elements of $z$ equal 0.
Starting with $S(x) = S_0(x \exp(-\gamma^t z))$, the hazard is given by, $h(x) = \frac {f(x)}{S(x)} = \frac {\frac{- dS(x)} { dx} } {S(x)}$.
$\frac{- dS(x)} { dx} = \frac { -d(S_0(x \exp(-\gamma^t z)) )} { dx} = f(x \exp(-\gamma^t z))\exp(-\gamma^t z) $.
We can then have: $h(x) = \frac{f( x \exp(-\gamma^t z))\exp(-\gamma^t z) } {S_0( x \exp(-\gamma^t z) )} = h_0(x \exp(-\gamma^t z)) \exp(-\gamma^t z)$
I am not too sure that $h_0(u) = f(u) / S_0(u) $. Will be happy for the feedback.
