Lognormal Frailty and Gaussian Random Effects

From Gaussian Random Effects to Lognormal Frailty

A frailty is a latent multiplicative factor in an event rate or hazard. A Gaussian random effect provides a convenient way to model such a factor: take the exponential of the Gaussian effect, so the resulting frailty is positive. The centering of that Gaussian effect matters. A zero-mean Gaussian effect gives a lognormal frailty whose mean is greater than one, whereas a particular negative shift produces a frailty with mean one.

The mean-one convention is useful when the baseline hazard is intended to describe the population-average frailty level at the start of follow-up. It does not make marginal survival equal to survival at the average frailty: survival is a nonlinear function of the frailty, so averaging conditional survival requires integration.

Definition: Let $W\sim N(0,1)$, and let $\sigma^2\geq 0$ be the variance of a Gaussian random effect. The mean-one lognormal frailty is $$ Z=\exp\left(\sigma W-\frac{\sigma^2}{2}\right). $$ Equivalently, if $U=\log Z$, then $U\sim N(-\sigma^2/2,\sigma^2)$. The frailty $Z$ is positive, has $E(Z)=1$, and has variance $\operatorname{Var}(Z)=e^{\sigma^2}-1$.

The variance parameter $\sigma^2$ is on the log-frailty scale; it is not the variance of $Z$. The identities follow from the Gaussian moment-generating function: if $U\sim N(\mu,v)$, then $E(e^{aU})=e^{a\mu+a^2v/2}$. Thus, for the centered choice $\mu=-\sigma^2/2$, $E(Z)=e^{-\sigma^2/2+\sigma^2/2}=1$ and $E(Z^2)=e^{-\sigma^2+2\sigma^2}=e^{\sigma^2}$, giving the stated variance.

Centering the Effect and Interpreting Its Scale

Suppose a Gaussian random intercept is written as $U\sim N(0,\sigma^2)$, with frailty $Z_0=e^U$. This is a valid lognormal multiplier, but it is not mean one: $E(Z_0)=e^{\sigma^2/2}$. To obtain mean one without changing the variance of the log frailty, subtract $\sigma^2/2$ from the random intercept: $Z=e^{U-\sigma^2/2}$. In a regression model, this is a shift in the intercept of the latent log multiplier, not a change to the random-effect variance.

Worked Example: Converting a Gaussian Random Intercept to Mean-One Frailty

Suppose a model uses a Gaussian random intercept $U\sim N(0,0.5)$ and the multiplicative effect $Z_0=e^U$. Find its mean and variance, then give the mean-one version with the same log-scale variance.

For a lognormal variable with log mean zero and log variance $0.5$, $$ E(Z_0)=e^{0+0.5/2}=e^{0.25}\approx 1.284, \qquad \operatorname{Var}(Z_0)=e^{1.0}-e^{0.5}\approx 1.070. $$

The corresponding mean-one frailty is $$ Z=e^{U-0.25}. $$ Its mean is $e^{-0.25+0.5/2}=1$, and its variance is $e^{0.5}-1\approx 0.6487$. The shift changes the mean of the multiplier but leaves the variance of its logarithm at $0.5$. A zero-mean random intercept is therefore not, by itself, evidence that the exponentiated frailty has mean one.

In a hazard model, one possible conditional specification is $\lambda(t\mid Z)=Z\lambda_0(t)$, where $\lambda_0(t)$ is a nonnegative baseline hazard. For a recurrent-event model, the analogous specification may be a conditional event intensity multiplied by $Z$. The survival calculation below applies to a conditional no-event probability of exponential cumulative form—for example, a first-event survival model, or a conditional Poisson event process. It is not a claim that every recurrent-event model has a first-event survival function of this form.

Model assumptions for the calculation: Conditional on $Z$, the event-free survival through time $t$ is $\exp\{-ZA(t)\}$, where $A(t)=\int_0^t\lambda_0(s)\,ds$ is finite and nonnegative. The frailty has the stated lognormal distribution, and the marginal quantity averages over that frailty distribution. If covariates are included, their values are held fixed in the conditional model and included in its cumulative hazard.

Marginal Survival as a Gaussian Expectation

Conditional survival is straightforward once the frailty is specified. Marginal survival averages it over the unobserved Gaussian effect. For the mean-one parameterization, write $W\sim N(0,1)$ and substitute $Z=\exp(\sigma W-\sigma^2/2)$. This yields an expectation over a standard normal variable; in general, the resulting integral has no elementary closed form.

S_{\mathrm{marg}}(t) =E\!\left[\exp\{-ZA(t)\}\right] =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\!\left[-A(t)e^{\sigma w-\sigma^2/2}\right]e^{-w^2/2}\,dw.

The same formula with $e^{\sigma w}$ in place of $e^{\sigma w-\sigma^2/2}$ applies to the uncentered frailty $Z_0=e^{\sigma W}$. It is important to use the parameterization that the model actually specifies; centering is not an optional numerical adjustment.

A convenient numerical method is three-node Gauss–Hermite quadrature for a standard normal expectation. It uses nodes $w=0,\pm\sqrt{3}$, with weights $2/3,1/6,1/6$, respectively. These weights sum to one and their weighted squared nodes equal one, as required to reproduce the mean and variance of a standard normal. For a function $g$, the approximation is

E\{g(W)\}\approx \frac{1}{6}g(-\sqrt{3})+\frac{2}{3}g(0)+\frac{1}{6}g(\sqrt{3}).

Consequently, for mean-one lognormal frailty, $$ S_{\mathrm{marg}}(t)\approx \frac{1}{6}e^{-A(t)e^{-\sqrt{3}\sigma-\sigma^2/2}} +\frac{2}{3}e^{-A(t)e^{-\sigma^2/2}} +\frac{1}{6}e^{-A(t)e^{\sqrt{3}\sigma-\sigma^2/2}}. $$ This is a numerical approximation, not an exact identity. The nodes are for $W$; they must be multiplied by $\sigma$ and then shifted by $-\sigma^2/2$ to obtain nodes for the mean-one log frailty.

Worked Numerical Integrations

Worked Example: Three-Node Approximation With Mean-One Frailty

Suppose $\sigma^2=0.5$ and the cumulative baseline hazard at time $t$ is $A(t)=0.7$. Approximate marginal survival using three-node Gauss–Hermite quadrature.

Here $\sigma=\sqrt{0.5}$, so the three log-frailty values, in order of negative, zero, and positive Gaussian nodes, are $-\sqrt{1.5}-0.25$, $-0.25$, and $\sqrt{1.5}-0.25$. Exponentiating gives frailty values approximately $0.2288,\ 0.7788,\ 2.6505$. The corresponding conditional survivals are $$ e^{-0.7(0.2288)}\approx 0.8520,\qquad e^{-0.7(0.7788)}\approx 0.5797,\qquad e^{-0.7(2.6505)}\approx 0.1564. $$

Apply the quadrature weights, pairing the smallest frailty with the negative node and the largest with the positive node: $$ S_{\mathrm{marg}}(t)\approx \frac{1}{6}(0.8520)+\frac{2}{3}(0.5797)+\frac{1}{6}(0.1564) =0.5545. $$ Thus the approximate marginal probability of remaining event-free through $t$ is $0.5545$, under the specified model and cumulative baseline hazard. It is not the conditional survival at frailty one, which would be $e^{-0.7}\approx0.4966$.

Worked Example: Why the Mean-One Shift Changes the Answer

Keep $\sigma^2=0.5$, but first use the uncentered multiplier $Z_0=e^U$, where $U\sim N(0,0.5)$, and set $A(t)=0.5$. Apply the same three-node approximation.

The log-frailty nodes are $U=-\sqrt{1.5},0,\sqrt{1.5}$, so the corresponding multipliers are approximately $0.2938,1.0000,3.4033$. The conditional survivals are approximately $$ e^{-0.5(0.2938)}=0.8634,\qquad e^{-0.5(1.0000)}=0.6065,\qquad e^{-0.5(3.4033)}=0.1824. $$

Weighting the three values gives $$ \frac{1}{6}(0.8634)+\frac{2}{3}(0.6065)+\frac{1}{6}(0.1824) \approx 0.5787. $$ This calculation is for the uncentered multiplier, whose mean is $e^{0.25}\approx1.284$. Replacing it by the mean-one multiplier changes all the frailty values and therefore changes the marginal survival. One cannot switch parameterizations while keeping the same numerical interpretation of the baseline hazard.

Worked Example: A Smaller Log-Frailty Variance

Now let $\sigma^2=0.25$, use mean-one frailty, and take $A(t)=0.5$. Here $\sigma=0.5$, so the log-frailty values at the three standard-normal quadrature nodes are $-\sqrt{3}(0.5)-0.125$, $-0.125$, and $\sqrt{3}(0.5)-0.125$. The frailties are approximately $0.3712,0.8825,2.098$.

Their conditional survivals are $$ e^{-0.5(0.3712)}\approx0.8306,\qquad e^{-0.5(0.8825)}\approx0.6432,\qquad e^{-0.5(2.098)}\approx0.3503. $$

Therefore, $$ S_{\mathrm{marg}}(t)\approx \frac{1}{6}(0.8306)+\frac{2}{3}(0.6432)+\frac{1}{6}(0.3503) \approx0.6256. $$ The values at the outer nodes differ substantially, but their weights are small. The result is an approximation to the distributional average, not an average of the three frailty values followed by one survival calculation.

Justification, Diagnostics, and Limitations

The marginal-survival formula follows directly from conditioning. Let $E_t$ denote the event that the individual remains event-free through $t$. By the law of total probability, $$ P(E_t)=E\{P(E_t\mid Z)\}. $$ Under the stated conditional model, $P(E_t\mid Z)=e^{-ZA(t)}$. Substituting the mean-one lognormal expression for $Z$ gives the Gaussian expectation and integral above. The quadrature formula then replaces that expectation by a weighted sum. Each step requires the specified conditional survival model and frailty distribution; quadrature does not repair a misspecified model.

Three-node quadrature is useful for hand calculations and may be adequate when the integrand is sufficiently smooth over the important part of the Gaussian distribution. It is not guaranteed to be accurate for every value of $A(t)$ and $\sigma^2$. In computation, compare the result with quadrature using more nodes or with a direct numerical evaluation of the one-dimensional integral. Agreement to the required precision is a practical check. For extreme parameter values, evaluate the conditional survivals and weighted sum with numerically stable methods.

Model diagnostics address a different question from quadrature accuracy. Check whether the assumed lognormal random-effect distribution and the conditional hazard or intensity structure are plausible for the application. A discrepancy between low- and higher-order integration can signal numerical error; a good agreement between them says nothing by itself about whether the lognormal distribution is appropriate. Frailty variance may also be difficult to distinguish from changes in the baseline hazard without adequate information, so estimates and their uncertainty should be interpreted in the context of the data and model.

Approach	What it provides	Important qualification
Mean-one lognormal frailty	A positive multiplier represented by an exponentiated Gaussian effect, with an interpretable mean-one scale.	Marginal survival generally requires numerical integration.
Uncentered Gaussian random effect	A familiar zero-mean Gaussian effect on the log scale.	Its exponent has mean $e^{\sigma^2/2}$, not one; baseline interpretation changes accordingly.
Gamma frailty	An alternative positive frailty distribution; its Laplace transform can simplify marginal calculations.	It imposes a different distributional shape and is not interchangeable with a lognormal frailty.

The practical choice between distributions depends on the scientific model and the information in the data, not only on whether an integral is convenient. A lognormal effect links directly to the commonly used Gaussian random-intercept formulation, but its marginal survival integral generally has no elementary closed form. Conversely, a convenient calculation under another frailty distribution does not establish that distribution as appropriate.

Common Mistakes

Assuming that a zero-mean Gaussian effect gives mean-one frailty. Exponentiation changes the mean: $E(e^U)=e^{\sigma^2/2}$ when $U\sim N(0,\sigma^2)$. Subtract $\sigma^2/2$ on the log scale to make the multiplier mean one.
Using the wrong quadrature nodes. For a standard normal expectation with three nodes, use $0,\pm\sqrt{3}$ and weights $2/3,1/6,1/6$. If the log-frailty variance is $\sigma^2$, scale those nodes by $\sigma$; for mean-one frailty, also subtract $\sigma^2/2$.
Averaging frailties before calculating survival. In general, $E\{e^{-ZA}\}\neq e^{-E(Z)A}$, because the exponential function is nonlinear. Average the conditional survival values instead.
Treating a quadrature approximation as exact. Verify numerical stability by increasing the number of nodes or evaluating the integral directly when precision matters.

Key takeaway: A mean-one lognormal frailty is the exponential of a Gaussian effect shifted by minus one-half its variance. Marginal survival is the expectation of conditional survival over that frailty distribution, and numerical integration must preserve both the centering and the Gaussian quadrature weights.

Check Your Understanding

Answer each question, then compare with the worked solutions below.

If $U\sim N(0,0.4)$, what is the mean of $e^U$, and what shift gives an exponentiated effect with mean one?
For mean-one lognormal frailty with $\sigma^2=0.4$, write the frailty at a standard-normal quadrature node $w$.
Why is $e^{-A(t)}$ generally not the marginal survival under mean-one frailty?
What practical check can assess whether three-node quadrature is adequate?

Solutions. (1) The mean is $e^{0.4/2}=e^{0.2}\approx1.221$. Use $Z=e^{U-0.2}$; then $E(Z)=e^{-0.2+0.4/2}=1$. (2) Since $\sigma=\sqrt{0.4}$, the frailty at node $w$ is $e^{\sqrt{0.4}w-0.2}$. (3) Marginal survival is $E\{e^{-ZA(t)}\}$, whereas $e^{-A(t)}$ is conditional survival at $Z=1$. Nonlinearity means these are generally unequal, even though $E(Z)=1$. (4) Recalculate with more quadrature nodes or directly evaluate the one-dimensional Gaussian integral, then check whether the result changes materially at the desired precision.