Frailty Effects in Parametric Survival NLME

Why Shared Frailty Changes Survival

People in the same household, treatment center, or other cluster may share unmeasured characteristics that affect their event risks. A shared frailty model represents that unobserved heterogeneity with a positive random multiplier on the hazard. Conditional on the multiplier, the survival model can retain a familiar parametric form. After averaging over frailty, however, both the survival curve and the hazard generally change: high-frailty members tend to experience events earlier, leaving a progressively more robust group among those still at risk.

The distinction between conditional and marginal survival is central. A conditional curve describes survival for a specified frailty value. A marginal curve averages over the frailty distribution in the target population. The latter is not obtained by simply setting frailty equal to its mean in the conditional model.

Definition: Let $j$ index clusters and $i$ individuals within a cluster. A shared frailty $Z_j$ is a positive random variable that multiplies the hazard for every member of cluster $j$. A common specification is $$ h_{ij}(t\mid Z_j,\mathbf{x}_{ij})=Z_j h_0(t)\exp(\boldsymbol{\beta}^{\mathsf T}\mathbf{x}_{ij}), $$ where $h_0(t)$ is the baseline hazard, $\mathbf{x}_{ij}$ is the covariate vector, and $\boldsymbol{\beta}$ is the coefficient vector. Frailty is commonly normalized so that $E(Z_j)=1$; its variance is denoted by $\theta=\operatorname{Var}(Z_j)$. Conditional on frailty and covariates, the cumulative hazard is $Z_j s_{ij}(t)$, where $s_{ij}(t)=H_0(t)\exp(\boldsymbol{\beta}^{\mathsf T}\mathbf{x}_{ij})$ and $H_0(t)=\int_0^t h_0(u)\,du$.

The model assumes that frailties are positive and independent across clusters, and that individuals are conditionally independent given their cluster’s frailty and their covariates. The frailty distribution is usually taken to be independent of measured covariates. The event-time distribution is specified parametrically through $h_0(t)$, and censoring must be noninformative conditional on the variables included in the likelihood. These conditions give the random multiplier a clear interpretation; they are not guaranteed by choosing a particular frailty distribution.

From Conditional to Marginal Survival

For a specified $Z_j=z$, an individual’s conditional survival is the exponential of the negative cumulative hazard. To obtain population survival, average this conditional survival over $Z_j$. Define the Laplace transform of the frailty distribution by $L_Z(s)=E[\exp(-sZ_j)]$, for $s\geq 0$. Then

S_{ij}(t\mid \mathbf{x}_{ij})=E_{Z_j}\!\left[\exp\{-Z_j s_{ij}(t)\}\right] =L_Z\!\left(s_{ij}(t)\right).

This expression is the key calculation: the frailty distribution determines the marginal survival curve through its Laplace transform. In general, $L_Z(s)$ is not equal to $\exp\{-sE(Z_j)\}$. Since $E(Z_j)=1$, replacing frailty by its mean would give $\exp(-s)$, which is the no-heterogeneity curve, not the mixture survival curve.

Worked Example: Gamma Frailty and Marginal Survival

Suppose the dimensionless cumulative hazard component at a specified time and covariate value is $s=0.75$, and frailty has a gamma distribution with mean one and variance $\theta=0.5$. For this parameterization, the gamma shape and rate are both $1/\theta$, and its Laplace transform is $L_Z(s)=(1+\theta s)^{-1/\theta}$.

The conditional survival for frailty $z$ is $\exp(-0.75z)$. For example, it is $\exp(-0.375)=0.6873$ at $z=0.5$, and $\exp(-1.125)=0.3247$ at $z=1.5$. Those values illustrate why survival differs across frailty levels; the marginal curve averages across the full gamma distribution, not just these two values.

Substituting into the Laplace transform gives $$ S(t)=(1+0.5\times0.75)^{-1/0.5} =(1.375)^{-2} =\frac{1}{1.890625} =0.5289. $$

Thus the modeled population survival at this time and covariate value is 0.5289. Setting $Z=E(Z)=1$ instead would give $\exp(-0.75)=0.4724$, which is not the marginal survival under gamma heterogeneity.

Common Frailty Distributions

The mean-one, positive frailty convention makes the baseline hazard the hazard for a unit frailty, while $\theta$ describes heterogeneity. Different distributions with the same mean and variance can still produce different marginal curves, especially away from time zero. Three commonly used choices are:

Frailty distribution	Mean-one parameterization	Laplace transform	Practical implication
Gamma	Shape and rate $1/\theta$	$(1+\theta s)^{-1/\theta}$	Closed-form marginal survival and hazard; widely used.
Lognormal	$\log Z\sim N(-\sigma^2/2,\sigma^2)$, with $\theta=e^{\sigma^2}-1$	$E[\exp(-sZ)]$, generally evaluated numerically	Allows a right-skewed frailty distribution; marginal survival typically has no elementary closed form.
Inverse Gaussian	Mean $1$, variance $\theta$	$\exp\{(1-\sqrt{1+2\theta s})/\theta\}$	Provides another positively skewed distribution with a closed-form transform.

The gamma model is convenient, but convenience is not evidence that gamma frailty describes a given population. A lognormal distribution can put relatively more weight in a long right tail, while the inverse-Gaussian transform gives a distinct marginal curve even at the same mean and variance. Distributional choice matters because the marginal curve depends on the full frailty distribution, not only its first two moments.

Worked Example: Comparing Gamma and Inverse-Gaussian Marginal Survival

Consider the same $s=1$ and $\theta=0.5$ under mean-one gamma and inverse-Gaussian frailty. Both distributions have mean frailty one and variance 0.5, but their transforms differ.

For gamma frailty, $$ S_{\mathrm{gamma}}=(1+0.5\times1)^{-1/0.5} =(1.5)^{-2} =\frac{1}{2.25} =0.4444. $$ For inverse-Gaussian frailty, $$ S_{\mathrm{IG}}=\exp\!\left\{\frac{1-\sqrt{1+2(0.5)(1)}}{0.5}\right\} =\exp\{2(1-\sqrt{2})\} =\exp(-0.8284) \approx0.4367. $$

The marginal survivals differ by about 0.0077 despite the matching means and variances. This comparison shows that the frailty distribution’s shape can affect predictions; it does not establish which distribution is appropriate for any particular data set.

Marginal Hazards and Survivor Selection

A conditional proportional-hazards model does not generally remain proportional after frailty is integrated out. Let $s(t)=H_0(t)\exp(\boldsymbol{\beta}^{\mathsf T}\mathbf{x})$ for a fixed covariate vector. Differentiating $S(t)=L_Z(s(t))$ gives the marginal hazard:

h_{\mathrm{marg}}(t\mid\mathbf{x}) =-\frac{d}{dt}\log L_Z(s(t)) =h_0(t)\exp(\boldsymbol{\beta}^{\mathsf T}\mathbf{x}) \frac{E[Z_j\exp\{-s(t)Z_j\}]}{E[\exp\{-s(t)Z_j\}]}.

The ratio in this expression is the mean frailty among individuals who have survived to time $t$, conditional on the covariates. This follows because the conditional density of frailty among survivors is proportional to $\exp\{-s(t)z\}$ times its population density. As time passes and cumulative hazard increases, high-frailty individuals are less represented among survivors. This depletion of susceptible individuals makes the marginal hazard a time-dependent quantity even when the conditional hazard has a simple proportional form.

Result: Under mean-one gamma frailty with variance $\theta>0$, $L_Z(s)=(1+\theta s)^{-1/\theta}$. Therefore $$ S_{\mathrm{marg}}(t\mid\mathbf{x})=(1+\theta s(t))^{-1/\theta}, \qquad h_{\mathrm{marg}}(t\mid\mathbf{x}) =\frac{h_0(t)\exp(\boldsymbol{\beta}^{\mathsf T}\mathbf{x})}{1+\theta s(t)}. $$ The marginal hazard is the conditional hazard multiplier $h_0(t)\exp(\boldsymbol{\beta}^{\mathsf T}\mathbf{x})$ times the mean frailty among survivors, $1/(1+\theta s(t))$.

Worked Example: A Declining Marginal Hazard Under an Exponential Baseline

Assume an exponential baseline hazard $h_0(t)=0.2$ per month, no covariate effect, and mean-one gamma frailty with $\theta=0.5$. The baseline cumulative hazard is $H_0(t)=0.2t$, so $s(t)=0.2t$.

The population survival is $$ S_{\mathrm{marg}}(t)=(1+0.5(0.2t))^{-2}=(1+0.1t)^{-2}. $$ At $t=5$ months, $s=1$, and $$ S_{\mathrm{marg}}(5)=(1+0.5\times1)^{-2}=1/2.25=0.4444. $$ The marginal hazard at that time is $$ h_{\mathrm{marg}}(5)=\frac{0.2}{1+0.5\times1} =\frac{0.2}{1.5} =0.1333\text{ per month}. $$

The conditional hazard for a member with $Z=1$ remains $0.2$ per month at all times. The lower marginal hazard at month 5 reflects selection among survivors: the mean frailty among survivors is $1/1.5=0.6667$, and $0.2\times0.6667=0.1333$ per month.

Shared Frailty Creates Within-Cluster Dependence

Members sharing $Z_j$ have independent event times only conditionally, given that frailty. Marginally, their event times are dependent. If two members have covariate-specific cumulative hazard components $s_1$ and $s_2$, their joint probability of surviving to their respective times is

P(T_{1j}>t_1,T_{2j}>t_2) =E[\exp\{-Z_j(s_1+s_2)\}] =L_Z(s_1+s_2).

If the two members instead had independent frailties, their joint survival would be $L_Z(s_1)L_Z(s_2)$. The distinction is the shared latent multiplier: observing that one cluster member survives provides information about the frailty, and therefore about the other member’s event risk.

Worked Example: Joint Survival With Shared Gamma Frailty

Suppose two members of one cluster have $s_1=0.40$ and $s_2=0.70$, and share mean-one gamma frailty with $\theta=0.5$. Their conditional joint survival given $Z$ is $\exp\{-Z(0.40+0.70)\}$. Integrating over their shared frailty gives $$ P(T_1>t_1,T_2>t_2) =(1+0.5(1.10))^{-2} =(1.55)^{-2} =\frac{1}{2.4025} =0.4162. $$

Their separate marginal survival probabilities are $$ S_1=(1+0.5(0.40))^{-2}=(1.20)^{-2}=0.6944, \qquad S_2=(1+0.5(0.70))^{-2}=(1.35)^{-2}=0.5487. $$ If their survival outcomes were independent, their joint probability would be $$ S_1S_2=(1.20)^{-2}(1.35)^{-2}\approx0.3810. $$ Thus the shared-frailty joint survival exceeds the independent-frailty product by approximately $0.4162-0.3810=0.0352$. This positive dependence arises because a relatively low frailty makes survival more likely for both members, while a relatively high frailty raises both hazards.

Likelihood, Interpretation, and Limitations

For an observed event time or censoring time $t_{ij}$, let $\delta_{ij}=1$ indicate an observed event and $\delta_{ij}=0$ indicate right censoring. Conditional on frailty, the event-time likelihood contribution is $h_{ij}(t_{ij}\mid Z_j)^{\delta_{ij}}S_{ij}(t_{ij}\mid Z_j)$. Under conditional independence within a cluster, the cluster likelihood integrates the product of these contributions over one shared frailty:

L_j =E_{Z_j}\!\left[ \prod_i h_{ij}(t_{ij}\mid Z_j)^{\delta_{ij}} S_{ij}(t_{ij}\mid Z_j) \right].

The integration is outside the product. Integrating each member’s contribution separately would represent independent member-specific frailties, not a shared cluster frailty. With gamma frailty, many such integrals can be evaluated in closed form; other distributions may require numerical integration.

Frailty coefficients and hazards require careful interpretation. A coefficient in $\boldsymbol{\beta}$ describes a covariate effect conditional on the shared frailty. Because the frailty distribution among survivors changes over time, a marginal hazard ratio between covariate groups need not equal $\exp(\beta_k)$ or remain constant over time. Marginal survival comparisons and conditional hazard comparisons answer different questions.

Frailty is especially useful when the scientific structure suggests common unobserved risk within clusters and when the data contain enough information to estimate that heterogeneity. It is not a generic remedy for every pattern of dependence. With only one event-time observation per cluster, frailty variance can be difficult to distinguish from the parametric baseline hazard shape. Even with repeated or clustered observations, a misspecified baseline or unmodeled covariate structure can be mistaken for frailty variation. Estimates may also be sensitive to the assumed frailty distribution, particularly when cluster sizes are small or event information is sparse.

A no-frailty model is the simpler alternative and corresponds to $\theta=0$, so $Z_j=1$ for all clusters. It implies no residual dependence from a shared latent multiplier. Independent individual frailties can describe individual heterogeneity, but they do not induce the same within-cluster dependence. Shared frailty is appropriate when the random effect is genuinely common to cluster members; these approaches are not interchangeable.

Check the scientific unit of sharing.
Specify which observations share a frailty, and confirm that the proposed clustering matches the source of unobserved risk.

State the conditional model.
Write the baseline hazard, covariate effect, frailty distribution, mean normalization, and conditional-independence assumptions before deriving marginal quantities.

Use the Laplace transform for population survival.
Substitute $s=H_0(t)\exp(\boldsymbol{\beta}^{\mathsf T}\mathbf{x})$ into $L_Z(s)$; do not substitute the mean frailty into the conditional survival formula.

Interpret the target quantity.
Distinguish conditional hazard effects from marginal survival and hazard, and examine whether the assumed distribution and baseline can be separated by the available cluster information.

Common pitfall: “The frailty has mean one, so marginal survival is just $\exp\{-s\}$.” This substitutes the mean into a nonlinear expectation, but in general $E[\exp(-sZ)]\ne\exp\{-sE(Z)\}$. Use the frailty Laplace transform; for gamma frailty, marginal survival is $(1+\theta s)^{-1/\theta}$, not $\exp(-s)$.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

For gamma frailty with mean one and variance $\theta$, what is the marginal survival when the covariate-specific cumulative hazard component is $s$?
Why does a shared frailty induce dependence between two cluster members’ survival outcomes?
With $h_0(t)=0.1$ per month, no covariate effect, gamma frailty variance $\theta=1$, and $t=4$ months, calculate marginal survival and marginal hazard.
State one reason a fitted frailty variance may be difficult to interpret when there is only one event-time observation per cluster.

Solutions. (1) The Laplace transform gives $S(t)=(1+\theta s)^{-1/\theta}$. (2) Both outcomes depend on the same random $Z_j$; conditional independence given $Z_j$ does not imply marginal independence after $Z_j$ is integrated out. (3) Here $s=H_0(4)=0.1\times4=0.4$. Therefore $S=(1+1\times0.4)^{-1}=1/1.4=0.7143$, and $h_{\mathrm{marg}}(4)=0.1/(1+0.4)=0.07143$ per month. (4) With one event-time observation per cluster, the data may provide limited information to distinguish latent frailty heterogeneity from the chosen baseline hazard shape.

Frailty distribution	Mean-one parameterization	Laplace transform	Practical implication
Gamma	Shape and rate \(1/\theta\)	\((1+\theta s)^{-1/\theta}\)	Closed-form marginal survival and hazard; widely used.
Lognormal	\(\log Z\sim N(-\sigma^2/2,\sigma^2)\), with \(\theta=e^{\sigma^2}-1\)	\(E[\exp(-sZ)]\), generally evaluated numerically	Allows a right-skewed frailty distribution; marginal survival typically has no elementary closed form.
Inverse Gaussian	Mean \(1\), variance \(\theta\)	\(\exp\{(1-\sqrt{1+2\theta s})/\theta\}\)	Provides another positively skewed distribution with a closed-form transform.