Risk measures

Discussion in 'SA3' started by indexo, Dec 7, 2020.

  1. indexo

    indexo Member

    Hi,

    A few questions on page 10 of CR Chapter 8:
    1. How does VaR fail as a coherent measure by failing the sub-additivity requirement?
    2. How does T-VaR and transformed loss measures work as a coherent measure i.e. how does it fit into the sub-additivity requirement?
    3. How does expected policyholder default not meet the linear homogeneity axiom?

    In the CR, the explanation to above is quite limited. Would really appreciate some help on this!
     
  2. CapitalActuary

    CapitalActuary Ton up Member

    It’s not clear what you’re asking. Are you after mathematical proof and/or counter examples as appropriate?

    VaR fails subadditivity because VaR of total risk can be more than the sum of VaRs on the individual sub-risks in some cases. However, sub-additivity does always hold for VaR on a class of distributions called “elliptical” distributions (which includes student t and normal). So if your individual risks are normally distributed, for example, the VaR of the total will be less than or equal to the sum of VaRs on each sub-risk. The difference between the sum of the sub-risk VaRs and the total risk VaR is the “diversification benefit” that arises from combining the sub-risks together.

    If you want an a example then this stack exchange has one: https://quant.stackexchange.com/questions/34121/non-subadditivity-of-var

    For another example just look at the Wikipedia page on coherent risk measures.

    For 2, there are 7 proofs of subadditivity of ES (equivalent to TVaR on continuous distributions) here: https://people.math.ethz.ch/~embrecht/ftp/Seven_Proofs.pdf

    If you know or understand all of this fully you understand more than any actuary I’ve met. It’s well beyond what you need to know for the exams, and also what is helpful to know for day-to-day actuarial practice.

    For 3, I can’t remember what that is without looking at the core reading. I’ll get back to you if I remember.
     
  3. Sherwin

    Sherwin Member

    Regarding the first question, I will explain it in a plain example. Let's make five scenarios with the equal probability.

    Scenario, A's Loss, B's Loss, (A+B)'s loss
    1, 100, 400, 500
    2, 200, 500, 700
    3, 300, 600, 900
    4, 400, 900, 1300
    5, 500, 700, 1200

    You can see that the second largest loss for A is 400 and the second largest loss for B is 700. However, for the total loss of A and B, the second largest loss is 1200, which is higher than 400+700. That broke the sub-additivity requirement.
     

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