The Weibull as the limiting distribution of the Burr distribution

Question

I often deal with "payout patterns" which are vectors of the cumulative percentage of a loss that has been paid over time. For example, for $t \in [0, 1, 2, 3, 4, 5]$ I may have $p_t = (5\%, 15\%, 60\%, 85\%, 98\%, 100\%)$. As these vectors have to be between $0$ and $1$, I tend to used smoothed versions obtained by finding the "closest" parametric cumulative distribution function, almost always minimizing the sum squared distance between the empirical value and the CDF at each observations (similar to finding the distribution with the minimum Cramer-von Mises criterion).

Two of the distributions I often use are the Burr and the Weibull. I use the common North American actuarial parametrization as brought in Klugman, Panjer, and Wilmott (1998). For clarity, in terms of the distribution functions, they are: $$ \begin{align} \Large F(x)_\textrm{Weibull} &= \LARGE 1 - e^{-\left(\frac{x}{\theta}\right)^\tau}\\ \Large F(x)_\textrm{Burr} &= \Large 1 - \left(\frac{1}{1 + \left(\frac{x}{\theta}\right)^\gamma}\right)^\alpha \end{align} $$

What I have seen many times when solving for the minimum distance, is that if the Burr's $\theta$ and $\alpha$ diverge to $\infty$, the Burr CDF approaches the Weibull and the Burr $\gamma$ is the Weibull $\tau$.

I have sketched out a framework for a proof, but I a do not believe it is rigorous. What I would appreciate is:

1. Corrections, comments, or any constructive criticism on whether or not this relationship can be formally proven, and
2. Whether there is any way to estimate the Weibull $\theta$ from the diverging Burr (which I doubt for reasons brought below).

Proof attempt

We have the two distributions brought above. The question can be stated as proving: $$ \lim_{\alpha, \theta \rightarrow \infty} \left(\frac{1}{1 + \left(\frac{x}{\theta}\right)^\gamma}\right)^\alpha \to \Large e^{-\left(\frac{x}{\theta}\right)^\tau} $$

Firstly, re-write the left side as: $$ \lim_{\alpha, \theta \rightarrow \infty}\left({1 + \frac{x^\gamma}{\theta^\gamma}}\right)^{-\alpha} $$ Now let $\xi = \theta^\gamma$. We now have $$ \lim_{\alpha, \xi \rightarrow \infty}\left({1 + \frac{x^\gamma}{\xi}}\right)^{-\alpha} $$ Now here is the weak part. As both $\alpha$ and $\xi$ are approaching $\infty$, replace both with $n$. Firstly, I'm not sure that is necessarily mathematically legal, and secondly, this is where question 2 probably fails, since notwithstanding that both $\alpha$ and $\xi (\theta^\gamma)$ diverge, they do so at different rates.

That being said, we now have $$ \lim_{n \rightarrow \infty}\left({1 + \frac{x^\gamma}{n}}\right)^{-n} $$

Taking the log and re-arranging, we get a L'Hopital condition of $$ \lim_{n \rightarrow \infty} \frac{\log\left[1 + \frac{x^\gamma}{n}\right]}{-\frac{1}{n}} $$ as both numerator and denominator approach 0.

Taking the individual derivatives of the numerator and denominator we get $$ \lim_{n \rightarrow \infty} \Large \frac{\left[\frac{1}{1 + \frac{x^\gamma}{n}}\right]\cdot\left({-x}^\gamma\right) n^{-2}}{n^{-2}} $$

The powers of $n$ cancel leaving us with $$ \lim_{n \rightarrow \infty} \Large \left[\frac{1}{1 + \frac{x^\gamma}{n}}\right]\cdot\left({-x}^\gamma\right) $$

which converges to $-x^\gamma$. Since we originally logged the formula, we actually have $$ \lim_{\alpha, \theta \rightarrow \infty} \left(\frac{1}{1 + \left(\frac{x}{\theta}\right)^\gamma}\right)^\alpha \to \Large e^{-x^\gamma} $$

which is very close to what we actually want. What is missing is the $\theta^\gamma$ in the denominator, realizing that this $\theta$ has nothing to do with that found in the Burr.

So, I would appreciate help in making this more rigorous, if possible, understanding any limitations, and any other opportunities to learn. Thank you.

Answer 1

There is some confusion about the role of $\theta$ in both distributions. If you're taking a limit as $\theta \to \infty$ for a Burr distribution, then how can you expect the result to remain a function of $\theta$? So this is the first problem you need to address by being clear about the number of parameters there are in each model and how they might be related between models.

So, suppose we have $X \sim {\rm Weibull}(\tau, \theta_1)$ and $Y \sim {\rm Burr}(\alpha, \gamma, \theta_2)$, with CDFs given by $$\begin{align*} F_X(x) &= 1 - \exp\bigl(-(x/\theta_1)^\tau\bigr), \\ F_Y(y) &= 1 - \bigl(1 + (y/\theta_2)^\gamma\bigr)^{-\alpha}. \end{align*}$$ Also recall that $$\lim_{n \to \infty} (1 + z/n)^{-n} = e^{-z}.$$ This suggests that to get the CDF of $Y$ to approach (in a pointwise sense) the CDF of $X$, we would need to take the limit of $F_Y$ as $\alpha \to \infty$ and $\color{red}{\boxed{\theta_2 = \alpha^{1/\tau} \theta_1}}$. If $\theta_2$ is allowed to approach infinity independently of $\alpha$, then of course the limit of $F_Y$ is indeterminate. Hence we require setting $$\color{green}{\boxed{Y \sim {\rm Burr}(\alpha, \tau, \alpha^{1/\tau} \theta_1)}}$$ from which we obtain $$ \begin{align*} \lim_{\alpha \to \infty} F_Y(y) &= 1 - \lim_{\alpha \to \infty} \biggl(1 + \Bigl(\frac{y}{\alpha^{1/\tau} \theta_1}\Bigr)^\tau\biggr)^{-\alpha} \\ &= 1 - \lim_{\alpha \to \infty} \biggl(1 + \frac{(y/\theta_1)^\tau}{\alpha} \biggr)^{-\alpha} \\ &= 1 - \exp\bigl(-(y/\theta_1)^\tau\bigr) \\ &= F_X(y). \end{align*}$$