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.

  • $\begingroup$ Leaving the result itself aside, I think the general approach is nonsensical. You're taking cumulated payments, and as far as I can tell, fitting curves to them as if they were independent (by OLS). This can easil lead to weird, even nonsensical outcomes. Individual payments by period are probably still correlated (due to effects like economic and social inflation, for example), but nothing like as strongly as cumulated payments. $\endgroup$
    – Glen_b
    Jun 13, 2014 at 3:54
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    $\begingroup$ I am using a fit for the purpose of smoothing, interpolation, and extrapolation. For example, sometimes we have annual payments/reserve amounts but need to estimate the quarterly values. Also, empirical observations tend to be 'jagged' which is more often noise than signal. The "fit" is a version of minimum distance, merely finding the parametric curve which is "closest" in a squared error sense to each of the 10, 20, or 30 years at times, of payment observations (Workers Compensation can take decades for an accident year to run off, for example). I'm not using regression. $\endgroup$
    – Avraham
    Jun 13, 2014 at 3:58
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    $\begingroup$ I added the backstory as I thought it would be interesting to the reader. Feel free to ignore why I encountered this phenomenon. The question remains, can my observation be rigorously proven or not. Thanks. $\endgroup$
    – Avraham
    Jun 13, 2014 at 14:04

1 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 $$ \begin{align*} \lim_{\alpha \to \infty} F_Y(y) &= 1 - \lim_{\alpha \to \infty} \biggl(1 + \Bigl(\frac{y}{\alpha^{1/\tau} \theta_1}\Bigr)^\tau\biggr)^{-\alpha} \\ &= 1 - \lim_{\alpha \to \infty} \biggl(1 + \frac{(y/\theta_1)^\tau}{\alpha} \biggr)^{-\alpha} \\ &= 1 - \exp\bigl(-(y/\theta_1)^\tau\bigr) \\ &= F_X(y). \end{align*}$$

  • $\begingroup$ Thank you. Can we arbitrarily set $\theta_2 = \alpha^{\frac{1}{\tau}}\theta_1$ since both $\alpha$ and $\theta$ approach infinity in the limit? I've tried looking at relationships between the diverging parameters and the Weibull $\theta$ that remains and was not able to find anything reasonable. Which is why I tried that "trick" with setting them both equal to $n$ since, in the limit, they both approach $+\infty$ anyway. $\endgroup$
    – Avraham
    Jun 16, 2014 at 14:33
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    $\begingroup$ What the above establishes is that there exists a particular subset/choice of Burr parameters for which the limiting distribution tends to a Weibull. It says that we require a specific relationship between the parameters of both distributions in order for this to hold. It will not hold if $\theta_2$ tends to infinity at a different rate than what is implied by the relationship $\theta_2 = \alpha^{1/\tau} \theta_1$. $\endgroup$
    – heropup
    Jun 16, 2014 at 15:21
  • $\begingroup$ Interesting. So when empirically, I find that when I apply a non-linear numerical optimizer to the squared difference between the Burr CDF using $\alpha, \gamma, \theta_2$ and that $\alpha$ and $\theta$ diverge, but the resulting CDF is point-equivalent to a Weibull where $\tau = \gamma$, perforce $\theta_1$ must equal the limit of $\alpha^\tau\theta_2$? Thanks! $\endgroup$
    – Avraham
    Jun 16, 2014 at 15:31
  • $\begingroup$ Can you comment on my last question (above), please? Thanks. $\endgroup$
    – Avraham
    Jun 19, 2014 at 6:08
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    $\begingroup$ No that is not correct. If $\theta_2 = \alpha^{1/\tau} \theta_1$, then $\theta_1 = \alpha^{-1/\tau} \theta_2$; this much is obvious because if $\tau > 0$, $\alpha \to \infty$ and $\theta_2 \to \infty$ implies $\alpha^\tau \theta_2 \to \infty$ which is absurd, as we require $\theta_1$ to be finite. But in regard to the relationship implied by such a convergence, I would guess that, yes, it implies that the Weibull scale parameter is the constant of proportionality where $\theta_2 \propto \alpha^{1/\tau}$. $\endgroup$
    – heropup
    Jun 19, 2014 at 6:42

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