I am doing research into the three parameter Generalized Pareto Distribution $$ f(x|a,b,c) = \frac 1 b\left(1+a\left(\frac{x-c}{b}\right)\right)^{\big(-1-\frac 1 a\big)} $$ for finding VaR and CVaR. $x$ is a vector of returns greater than or equal to $c$. The paper Parameter Estimation for 3-parameter gpd by the principle of maximum entropy by Singh and Guo (paper) provides MLE estimators in equations (45) and (46). Given a known value of $c$, how do I calculate $a$ and $b$ using the provided estimators?

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    $\begingroup$ No precise reference here or URL. No explanation of the parameterisation; we need to see a density, distribution or quantile function formula. You seem to be referring to some $\xi$ and $\sigma$; what is the third parameter? Is $n$ sample size? So needs more documentation and explanation, please. $\endgroup$ – Nick Cox Oct 30 '15 at 19:52
  • $\begingroup$ Question updated. $\endgroup$ – salisboss Oct 30 '15 at 22:21
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    $\begingroup$ 1. the RHS's of eqns 45 and 46 are not MLEs. They're the RHS of equations you need to solve to get MLEs, and they ARE set equal to what's given on the LHS of those equations; you just need a third equation for $c$. $\:$ 2. They explain why there isn't one for $c$; they then take $c$ to be the smallest observation in the paragraph immediately below eqn 46, giving the 3 equations you need. (Their comment about the likelihood being unbounded seems to ignore the fact that the actual likelihood must drop to zero when $c$ is above the smallest data value, but they still did the right thing with c) $\endgroup$ – Glen_b Oct 31 '15 at 0:08
  • $\begingroup$ I understand that the paper introduces. Another good thing about the paper is the inclusion of the MLE estimators. I just don't what to do with the estimators or how to use them. I have already selected a value for the c parameter in my research. $\endgroup$ – salisboss Oct 31 '15 at 0:14
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    $\begingroup$ The correct MLE for $c$ in a single sample is the smallest observation. As long as your "selected" $c$ is smaller than all your data the other two equations should work. You should edit to explain your real question ... "given a value for $c$, how do I calculate the estimates of $a$ and $b$ from those two equations?" $\endgroup$ – Glen_b Oct 31 '15 at 0:17

Later edit: I give what seems to be a better solution here.

Note that the paper uses a different parameterization from the form given in the question. As Yves noted in comments, it uses $-a$ in place of your $a$ (both are common parameterizations; the only difficulty may be when it is unclear which parameterization is being used). If you convert answers back to your parameterization you'll have to make the corresponding change.

The paper says:

The MLE estimators can be expressed as:

$\sum_{i=1}^n \frac{(x_i-c)/b}{1-a(x_i-c)/b}=\frac{n}{1-a}\qquad\qquad\qquad$ (45)

$\sum_{i=1}^n \ln[1-a(x_i-c)/b]=-na\qquad\,$ (46)

[...] Clearly the likelihood function is maximum with respect to $c$ when $c=x_1$.

In fact there's some minor issues with the exposition; at the point where they set the derivative of the log-likelihood to zero, they're no longer dealing with $a,b$ and $c$ but with the estimators, $\hat{a},\hat{b}$ and $\hat{c}$. So they should really have:

i. $\hat{c} = x_{(1)}$ (at least it didn't appear that the sample was ordered until at that point they suddenly declare $x_1$ to be smallest; better to be explicit)

ii. Given the ML estimate of $c$, the parameters $a$ and $b$ are then estimated by simultaneously solving these two equations:

$\sum_{i=1}^n \frac{(x_i-\hat{c})/\hat{b}}{1-\hat{a}(x_i-\hat{c})/b}=\frac{n}{1-\hat{a}}\qquad\qquad\qquad$ (45)

$\sum_{i=1}^n \ln[1-\hat{a}(x_i-\hat{c})/\hat{b}]=-n\hat{a}\qquad$ (46)

for $\hat{a}$ and $\hat{b}$.

The idea is that given $\hat{c}$, you find $\hat{a}$ and $\hat{b}$ that make equations (45) and (46) true.

If you were to solve this pair of nonlinear equations simultaneously, generally you'd need some iterative scheme set up* that you can update the estimates numerically until (45) and (46) are very close to true.

*(starting with some reasonable guesses, such as method of moments or quantile-based estimates or by assuming $a=0$ and using the resulting ML estimate from an exponential for $\hat{b}$)

It's certainly possible to do so... however, most people would back up a step; rather than taking the derivative and setting it equal to zero and looking for an iterative scheme to solve the equations for $\hat{a}$ and $\hat{b}$, we can simply employ optimization methods to minimize the negative of the log-likelihood function and take as our parameter estimates the values of the parameters that give that minimum. That's what's usually done for the generalized Pareto.

This would again start with some reasonable guesses for $\hat{a}$ and $\hat{b}$ and iterate to reduce $-\ell=-\log\mathcal{L}$ until some minimum was effectively reached. One benefit of doing so is that once you've found your minimum, it's easier to get second derivative estimates out and so get asymptotic standard errors for the estimates.

[In practice with ML, since the likelihood function might not always be unimodal, it's often a good idea to evaluate it over a grid of plausible values to identify whether there are multiple local minima in $-\ell$ or other issues that might be relevant.]

  • $\begingroup$ Thanks for your help. Now to implement the minimization. $\endgroup$ – salisboss Oct 31 '15 at 2:53
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    $\begingroup$ You should be aware that unless some constraints on the shape parameter $a$ are used, the maximal likelihood is $\infty$ and is always reached for any $a < -1$ and b such that $1 + a x_{(n)}/b = 0$. This is most often ignored in programs because $-\log(1 + a x_{(n)}/b)$ remains numerically small (say $\leq 40$ ) while it should normally go to $\infty$ with an infinite precision. $\endgroup$ – Yves Dec 2 '15 at 9:47
  • $\begingroup$ @Glen_b There seems to be a change of notation $a \rightarrow -a$ from the question to your answer (each of the two choices being frequently used). Maybe you could warn about this at the begining of your answer? $\endgroup$ – Yves Oct 12 '18 at 12:45
  • $\begingroup$ Thanks @Yves - you're quite right about there being a mismatch, but the mismatch is really contained in the question. The big problem is that the change is not going to be obvious to readers who don't look at the paper the OP refers to; ideally that discrepancy between the form OP gives and the one OP's reference uses should be explained in the question. My answer uses the notation from OP's reference. In the absence of it being noted in the question, this discrepancy should be explicitly pointed out at the top of my answer, as you suggest. I will edit now. $\endgroup$ – Glen_b Oct 13 '18 at 1:17
  • $\begingroup$ That is clear and neat. $\endgroup$ – Yves Oct 13 '18 at 7:01

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