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Verification of internal risk measure estimates

  • Mark H. A. Davis EMAIL logo
Veröffentlicht/Copyright: 14. Januar 2016
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Abstract

This paper concerns sequential computation of risk measures for financial data and asks how, given a risk measurement procedure, we can tell whether the answers it produces are ‘correct’. We draw the distinction between ‘external’ and ‘internal’ risk measures and concentrate on the latter, where we observe data in real time, make predictions and observe outcomes. It is argued that evaluation of such procedures is best addressed from the point of view of probability forecasting or Dawid’s theory of ‘prequential statistics’ [12]. We introduce a concept of ‘calibration’ of a risk measure in a dynamic setting, following the precepts of Dawid’s weak and strong prequential principles, and examine its application to quantile forecasting (VaR – value at risk) and to mean estimation (applicable to CVaR – expected shortfall). The relationship between these ideas and ‘elicitability’ [24] is examined. We show in particular that VaR has special properties not shared by any other risk measure. Turning to CVaR we argue that its main deficiency is the unquantifiable tail dependence of estimators. In a final section we show that a simple data-driven feedback algorithm can produce VaR estimates on financial data that easily pass both the consistency test and a further newly-introduced statistical test for independence of a binary sequence.

MSC 2010: 62A01; 91B30; 91G70

A The Markov chain model

In model μ,θ the transition probabilities of the chain are as shown in Table 4, where f=(1-μ)/μ1. The table also indicates the notation ni, i=1,2,3,4 we use for the number of occurrences of the four pairs 00,11,01,10 in a sample of size n. It should be clear that n3 and n4 play no real role in the problem, since algebraically it must be the case that |n3-n4|1, so for a large sample n3n412(n-n1-n2).

Table 4

Markov chain transition probabilities, f=(1-μ)/μ. The sample size is n=n1+n2+n3+n4.

xk-1xkpθ(xk|xk-1)pμ(xk|xk-1)# in sample
001-θ1-μn1
111-θfμn2
01θμn3
10θf1-μn4

For any k the probability mass distribution of Yk is m(x)=1-μ+(2μ-1)x, x=0,1, while for n>0 the distribution of (Y0,,Yn) is

pnθ(x0,,xn)=m(x0)k=1npθ(xk|xk-1).

When θ=μ the Yk are i.i.d. with joint distribution pnμ(x0,,xn)=k=0nm(xk) so, referring to Table 4, the likelihood ratio LRn=pnθ/pnμ is given by

LRnθ(x0,,xn)=(1-θ)n1(1-θf)n2θn3(θf)n4(1-μ)-(n1+n4)μ-(n3+n2)
={(1-θ)n1(1-θf)n2θ(n3+n4)}{fn4(1-μ)-(n1+n4)μ-(n3+n2)}
={(1-θ)n1(1-θf)n2θn-n1-n2}{fn4(1-μ)-(n1+n4)μ-(n3+n2)}

and of course LRnμ1. The log likelihood ratio is therefore

LLRθ(n1,,n4)=Lθ(n1,n2)+M(n1,,n4)

where

(A.1)Lθ(n1,n2)=n1log(1-θ)+n2log(1-θf)+(n-n1-n2)logθ

and M=LLRθ-Lθ does not depend on θ.

Proposition A.1

For β12, the maximum likelihood estimator is given by

(A.2)θ^(n¯1,n¯2)=12f(n¯-2+fn¯-1-(f-c1)2+4f(c1-c2))

where n¯-i=(n-ni)/n=1-n¯i.

Proof.

To compute the maximum likelihood estimate we maximise Lθ over θ. We have

Lθθ=-n1(1-θ)-fn2(1-θf)+n-n1-n2θ=Q(θ)θ(1-θ)(1-θf)

where

(A.3)Q(θ)=fnθ2-(n-n2+(n-n1)f)θ+(n-n1-n2).

The discriminant of Q is

D=(n-n2+(n-n1)f)2-4fn(n-n1-n2)=n2[(f+c1)2-4fc2],

where

c1=1-n2+fn1n,c2=1-n1+n2n.

Under our standing assumption μ12 we have f1 and hence c1c2. Now D can be expressed as

D=n2[(f-c1)2+4f(c1-c2)],

showing that D0 whatever the values of n1,n2. Taking into account that Q(0), Q(1/f)>0 and Q(1)<0 we easily see that Q has a root in each of the intervals (0,1), (1,1/f) so the maximising θ^(0,1) is the smaller of the two roots, which is given by (A.2). ∎

Proposition A.2

In any model Hμ,θ0 with μ[12,1] and θ0[0,1], the estimator θ^ is consistent, that is, θ^(n¯1,n¯2)θ0 a.s. as n.

Proof.

Associated to the chain Yk is the 4-state Markov chain Yk, k=1,2, where Yk takes the values 1,2,3,4 respectively when (Yk-1,Yk)=(0,0),(1,1),(0,1),(1,0). The transition matrix for this chain is

𝐏=[1-θ00θ000θ001-θ00θ001-θ01-θ00θ00].

Consider first the case θ0(0,1). Then the chain Yk is irreducible and recurrent and consequently has a unique stationary distribution 𝐦 characterised by the property that 𝐦(𝐈-𝐏)=0, where 𝐈 is the 4×4 identity matrix. This system of equations is readily solved to give

𝐦=[(1-θ0)(1-μ)μ-(1-μ)θ0θ0(1-μ)θ0(1-μ)],

where we have substituted for θ0 from (5.4). The numbers ni introduced above are simply the numbers of visits to state i by the chain Y in a sample of length n. Since Y is recurrent,

(A.4)limnnin=𝐦ia.s.,i=1,2,3,4.

The quadratic form Q of (A.3) can be written as

1nQ(θ)=fθ2-(1-n¯2+(1-n¯1)f)θ+(1-n¯1-n¯2).

If we substitute n¯i=𝐦i, i=1,2 we find Q(θ0)=0 and hence that θ^(𝐦1,𝐦2)=θ0. Now θ^(n¯1,n¯2) is a continuous function of the two parameters, and hence in view of (A.4) we have limnθ^(n¯1,n¯2)=θ0 a.s.

We now consider the cases θ0=0,1. When θ0=0, Yk=Y0 for all k, so either n1=n, n2=0 or n1=0, n2=n giving, from (A.1), values of Lθ equal to nlog(1-θ) or nlog(1-θf) respectively. In either case, Lθ is maximised at θ=0=θ0.

The case θ0=1 is a little more tricky. Here Yk-1=0 implies Yk=1, so n10. The sample path consists of strings of ones separated by single zeros. The probability of flipping from 1 to 0 is f, so the mean length of a string of ones is 1/f. Each flip from 1 to 0 and back adds 1 to n3 and to n4, and each string of ones of length m adds m-1 to n2. So the mean growth rates in n3+n4 and in n2 are in the ratio 2:(1/f)-1=(2μ-1)/(1-μ), implying that, loosely speaking, the fraction of the time spent growing n2 is ((2μ-1)/(1-μ))/((2μ-1)/(1-μ)+2)=2μ-1. We conclude that

limnn¯2=2μ-1a.s.

At the limiting value,

1nLθ=(2μ-1)log(1-θf)+2(1-μ)logθ

and the derivative with respect to θ is

-μ(1-f)f1-θf+2(1-μ)θ.

This is equal to + at θ=0 and is finite and decreasing for θ>0. Its value at θ=1 is 1-μ>0, and we conclude that the maximum occurs at θ=1. A simple continuity argument now shows that

limnθ^(0,n¯2)=1=q0a.s.

Acknowledgements

The author thanks Paul Embrechts for introducing him to this topic, Phil Dawid for discussions of prequential statistics, and Johanna Ziegel for extremely helpful comments and advice. He is also grateful to two anonymous referees for their careful reading of the paper and many constructive suggestions for improvements. Some of this work was carried out at the Hausdorff Research Institute for Mathematics at the University of Bonn, where the author was a visitor during the Trimester Program Stochastic Dynamics in Economics and Finance in August 2013. Discussions there have been very helpful in clarifying the content of this paper.

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Received: 2015-3-15
Revised: 2015-9-17
Accepted: 2015-11-18
Published Online: 2016-1-14
Published in Print: 2016-12-1

© 2016 by De Gruyter

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