Testing the independence of violations, Section 8.3.2 p. 155-6 Financial risk forecasting errata

Testing the independence of violations, Section 8.3.2 p. 155-6

December 27, 2017

My FM320 students Hongshen Chen, Yida Li and Yanfei Zhou pointed out that the discussion in Section 8.3.2 could be more clear, so I repeat the relevant parts of the section here with more clarifications.

We need to calculate the probabilities of two consecutive violations, p11, as well the probability of a violation, if there was no violation on the previous day, i.e. p01. More generally, where i and j are either 0 or 1: pij=Pr(ηt=j|ηt1=i). The violation process can be represented as a Markov chain with two states, so the first order transition probability matrix is defined as: Π1=(1p01p01 1p11p11). The likelihood function is: (8.5)L1(Π1)=(1p01)v00p01v01(1p11)v10p11v11 where vij is the number of observations where j follows i.

The maximum likelihood (ML) estimates are obtained by maximizing the likelihood function which is simple since the parameters are the ratios of the counts of the outcomes: Π^1=(v00v00+v01v01v00+v01v10v10+v11v11v10+v11). Under the null hypothesis of no clustering, the probability of a violation tomorrow does not depend on today being a violation, then \(p_{01}=p_{11}=p\) and the transition matrix is simply: Π2=(1pp1pp) and the ML estimate is: p^=v01+v11v00+v10+v01+v11. so Π^2=(1p^p^1p^p^)

The likelihood function then is (8.6)L2(Π2)=(1p)v00+v10pv01+v11.

Note in (8.6) we impose independence but do not in (8.5). Replace the \(\Pi\) by the estimated numbers, \(\hat{\Pi}\). The LR test is then: LR=2(logL1(Π^1)logL2(Π^2))asymptoticχ(1)2.


Example 4.4
Sequential moments


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