Conditional Hazard Function in Frailty Models

Question

Let $\Lambda>0$ be an unobservable frailty random variable and let $X\sim f_X,F_X$ (PDF and CDF respectively), then we have: $$ h_{X|\Lambda}(x|\lambda)=\lambda a(x) $$ where $h_{X|\Lambda}(\cdot)$ is the conditional hazard function. I was wondering what allows this property, we do not know how $\lambda$ is used in $h_{X|\Lambda}(\cdot)$, that is, we do not know whether or not $\lambda$ is so easily factor from $h_{X|\Lambda}(\cdot)$. Also, how does this property lead to $$ \bar F_{X|\Lambda}(x|\Lambda=\lambda)=[\bar F(x)]^\lambda $$ where $\bar F_X(\cdot)$ is the survival function of X, that is, $\bar F_X(x)=1-F_X(x)$.

Answer 1

The conditional hazard relationship you cite is one definition of a frailty model, which is a generalization of the proportional hazards model. Thus, the conditional hazard $h_{X \mid \Lambda}(x \mid \lambda)$ is by construction said to represent a frailty model if it is proportional to a baseline hazard function $a(x)$, where $\Lambda$ is an unobserved random variable representing heterogeneous risks among the population. Consequently, there is nothing to prove in this statement.

Regarding the second equation, we recall that hazard equals density divided by survival; i.e., $$h_{X \mid \Lambda}(x \mid \lambda) = \frac{f_{X \mid \Lambda}(x \mid \lambda)}{\bar F_{X \mid \Lambda}(x \mid \lambda)} = \frac{-\bar F'_{X \mid \Lambda}(x \mid \lambda)}{\bar F_{X \mid \Lambda}(x \mid \lambda)} = -\frac{d}{dx}\left[\log \bar F_{X \mid \Lambda}(x \mid \lambda) \right].$$ But from the definition, the LHS is simply $\lambda a(x)$, thus integration immediately yields $$\bar F_{X \mid \Lambda}(x \mid \lambda) = \exp\left( -\lambda \int_{t = t_0}^x a(t) \, dt \right) = e^{-\lambda A(x)},$$ where $A(x) = \int_{t = t_0}^x a(t) \, dt$ for a suitably chosen $t_0$ is the cumulative baseline hazard. In turn, the unconditional survival is $$\bar F_{X}(x) = \int_{\lambda = 0}^\infty \bar F_{X \mid \Lambda}(x \mid \lambda) f_{\Lambda}(\lambda) \, d\lambda = \int_{\lambda = 0}^\infty e^{-\lambda A(x)} f_{\Lambda}(\lambda) \, d\lambda = M_\Lambda(-A(x)),$$ the moment-generating function of $\Lambda$ at $-A(x)$. At this point, we can see that we cannot establish a relationship between the conditional and unconditional survivals without further assumptions, since it is not generally true that $$e^{-\lambda A(x)} = (M_\Lambda(-A(x)))^\lambda.$$ One would need to state something about $M$ or $A$ or both. Indeed, the above would imply $e^{-A(x)} = M_\Lambda(-A(x))$, or $M_\Lambda(t) = e^t$, thus $\Lambda$ would be $1$ almost surely. You may wish to check if there has been a typographical error.