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Frailty models can be found in the field of survival analysis, particularly in the context of analyzing lifetime distributions. These models are used to account for the uncertainty associated with the hazard rate of an event or outcome. They are a type of mixture distribution that combines multiple components to generate new distributions. We can introduce a frailty random variable $\lambda$, which we will use to quantify the uncertainty over the hazard rate.

Specifically, let's consider the conditioned hazard function $h_{X|\Lambda}(x|\lambda)$ which equals to $\lambda a(x)$ where $a(x)$ is the baseline hazard function. We can also define the incomplete baseline hazard accumulation function $A(x) = \int_0^x a(t)dt$. Then, the survival function $S_{X|\Lambda}(x|\lambda)$ is given by $S_{X|\Lambda}(x|\lambda) = \exp(-A(x)\lambda)$. And the marginal survival function $S_X(x)$ is given by $S_X(x) = E_{\Lambda}[S_{X|\Lambda}(x|\lambda)] = E_{\Lambda}[\exp(-A(x)\lambda)]$, which is the moment generating function of $\Lambda$ evaluated at $-A(x)$.

When $a(x) = \gamma x^{\gamma-1}$, then $A(x) = x^{\gamma}$, our conditional probability will be $S_{X|\Lambda}(x|\lambda) = \exp(-x^{\gamma}\lambda) = \exp\left(- \left(\frac{x}{\lambda^{-1/\gamma}} \right)^{\gamma} \right)$ . This is the Weibull distribution.

In particular when $\gamma =1$, then we will have $S_{X|\Lambda}(x|\lambda) = \exp(-x\lambda)$, which is an exponential distribution.

Now if we have a Weibull distribution as our condition probability and lambda follows gamma distribution, we will have a Burr distribution as our marginal probability. We can show it by using the following.

First the moment generating function of a gamma distribution is the following:

$$M_{\Lambda}(-t) = \int_0^{\infty} \exp(-x t) \frac{x^{a-1}\exp(-x/\theta)}{\Gamma(a)\theta^a} dx $$

$$M_{\Lambda}(-t) = \frac{1}{\Gamma(a)\theta^a} \int_0^{\infty} x^{a-1}\exp(-x t-x/\theta) dx \\ =\frac{1}{\Gamma(a)\theta^a} \left(\frac{\theta}{\theta t+1}\right)^{a}\int_0^{\infty} \left( \left(\frac{\theta t+1}{\theta}\right)x\right)^{a-1}\exp\left(- \left(\frac{\theta t+1}{\theta}\right)x\right) d\left(\left(\frac{\theta t+1}{\theta}\right)x\right) $$

Where we can replace $\left(\frac{\theta t+1}{\theta}\right)x$ with $u$, so we will have $$M_{\Lambda}(-t) =\frac{1}{\Gamma(a)\theta^a} \left(\frac{\theta}{\theta t+1}\right)^{a}\int_0^{\infty} u^{a-1}\exp\left(- u\right) du = \left(\frac{1}{\theta t+1}\right)^{a} $$

Woops, so the moment generating function for gamma distribution is the following;

$$M_{\Lambda}(t) = \left(\frac{1}{1-\theta t}\right)^{a} $$

And now we also know that the marginal probability follows the following:

$$S_X(x) = M_{\Lambda}(-A(x)) $$

In this case, we have $A(x) = x^{\gamma}$, so by substitution, we will have

$$S_X(x) = \left(\frac{1}{1+\theta x^{\gamma}}\right)^{a}$$ which is a Burr distribution.

If we set $\gamma$ to be one, we will have $$S_X(x) = \left(\frac{1}{1+\theta x}\right)^{a}$$,

which, in turns, is a Pareto distribution