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(This article was first published on <strong><a href="https://statcompute.wordpress.com/2016/12/11/estimate-regression-with-type-i-pareto-response/"> S+/R – Yet Another Blog in Statistical Computing</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p>The Type-I Pareto distribution has a probability function shown as below<br>
<strong><br>
f(y; a, k) = k * (a ^ k) / (y ^ (k + 1))<br>
</strong><br>
In the formulation, the scale parameter <strong>0 &lt; a &lt; y</strong> and the shape parameter <strong>k > 1 </strong>. </p>
<p>The positive lower bound of Type-I Pareto distribution is particularly appealing in modeling the severity measure in that there is usually a reporting threshold for operational loss events. For instance, the reporting threshold of ABA operational risk consortium data is $10,000 and any loss event below the threshold value would be not reported, which might add the complexity in the severity model estimation. </p>
<p>In practice, instead of modeling the severity measure directly, we might model the shifted response <strong> y` = severity – threshold </strong> to accommodate the threshold value such that the supporting domain of y` could start from 0 and that the Gamma, Inverse Gaussian, or Lognormal regression can still be applicable. However, under the distributional assumption of Type-I Pareto with a known lower end, we do not need to shift the severity measure anymore but model it directly based on the probability function. </p>
<p>Below is the R code snippet showing how to estimate a regression model for the Pareto response with the lower bound <strong> a = 2 </strong> by using the <strong>VGAM</strong> package. </p>
<pre>
library(VGAM)
set.seed(2017)
n &lt;- 200
a &lt;- 2
x &lt;- runif(n)
k &lt;- exp(1 + 5 * x)
pdata &lt;- data.frame(y = rpareto(n = n, scale = a, shape = k), x = x)
fit &lt;- vglm(y ~ x, paretoff(scale = a), data = pdata, trace = TRUE)
summary(fit)
# Coefficients:
#             Estimate Std. Error z value Pr(>|z|)
# (Intercept)   1.0322     0.1363   7.574 3.61e-14 ***
# x             4.9815     0.2463  20.229  &lt; 2e-16 ***
AIC(fit)
#  -644.458
BIC(fit)
#  -637.8614
</pre>
<p>The SAS code below estimating the Type-I Pareto regression provides almost identical model estimation. </p>
<pre>
proc nlmixed data = pdata;
  parms b0 = 0.1 b1 = 0.1;
  k = exp(b0 + b1 * x);
  a = 2;
  lh = k * (a ** k) / (y ** (k + 1));
  ll = log(lh);
  model y ~ general(ll);
run;
/*
Fit Statistics
-2 Log Likelihood               -648.5
AIC (smaller is better)         -644.5
AICC (smaller is better)        -644.4
BIC (smaller is better)         -637.9

Parameter Estimate   Standard   DF    t Value   Pr > |t|
                     Error 
b0        1.0322     0.1385     200    7.45     &lt;.0001 	
b1        4.9815     0.2518     200   19.78     &lt;.0001 	
*/
</pre>
<p>At last, it is worth pointing out that the conditional mean of Type-I Pareto response is not equal to <strong> exp(x * beta) </strong> but <strong> a * k / (k – 1) </strong> with <strong> k = exp(x * beta) </strong>. Therefore, the conditional mean only exists when <strong> k > 1 </strong>, which might cause numerical issues in the model estimation. </p>
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