October 2011 Q8

Discussion in 'SP9' started by Jasmin Bedi, Sep 5, 2020.

  1. Jasmin Bedi

    Jasmin Bedi Member

    Hello, does any one have breakdown of the calculation for VaR and TVaR in October 2011 exam's Q8 that they can share please? Examiner's report is not so detailed.
     
  2. Anna Bishop

    Anna Bishop ActEd Tutor Staff Member

    Hi Jasmin - have you by any chance been on the ActEd tutorial for SP9? This question is in the Day 1 handout and there is a detailed solution. Let me know if not and I will post something more detailed here for you.
    Anna
     
  3. Jasmin Bedi

    Jasmin Bedi Member

    Thanks Anna, found it :)
     
    Anna Bishop likes this.
  4. Dar_Shan0209

    Dar_Shan0209 Ton up Member

    Hi Anna,
    Can you help me please with this question on the calculation of both VaR and TVaR?
    Thanks
     
  5. Anna Bishop

    Anna Bishop ActEd Tutor Staff Member

    HI Darshan

    With VaR questions, I find it helpful to start by writing down the definition from Sweeting:

    VaR(97.5%) = inf {l: P(L>l) ≤ 0.025}

    (Note 'inf' means we are looking for the smallest loss that satisfies the criteria in the curly brackets.)

    Then consider the distribution for the losses and work from there.

    Portfolio A

    The distribution of losses, L, is:
    • £2m prob 0.02 (if the bond defaults, we lose 20 * £100K = £2m)
    • –£100K prob 0.98 (if the bond doesn't default, we gain 20 * £5K = £100K, or a loss of –£100K)
    Then go through the values of the losses in turn to see which meets the criteria in the definition:
    • if l = £2m, then P(L>l) = 0 ≤ 0.025 and the criteria is met
    • if l = –£100K, then P(L>l) = 0.02 ≤ 0.025 and the criteria is met
    (Note that the inequality is strictly greater than in the probability.)

    Of these two bullets, the smallest loss that satisfies the criteria is l = –£100K.

    ie VaR(97.5%) = –£100K.

    Since VaR is expressed as a 'loss', this result is equivalent to a gain of £100K.
    We can interpret the result by saying there is a 97.5% chance that profits will be at least as great as £100K.

    Portfolio B

    The distribution of losses, L, is:
    • –£100K prob 0.6676 (if no bonds default, we gain 20 * £5K = £100K, or a loss of –£100K)
    • £5K prob 0.2725 (if one bond defaults, we gain 19 * £5K – 1* £100K = –£5K, or a loss of £5K)
    • £110K prob 0.0528 (if two bonds default, we gain 18 * £5K – 2* £100K = –£110K, or a loss of £110K)
    • £215K prob 0.0065 ...
    • £320K prob 0.0006 ...
    • £X prob 0
    Then go through the values of the losses in turn to see which meets the criteria in the definition:
    • if l = –£100K, then P(L>l) = 0.3324 > 0.025 and the criteria is NOT met
    • if l = £5K, then P(L>l) = 0.0599 > 0.025 and the criteria is NOT met
    • if l = £110K, then P(L>l) = 0.0071 ≤ 0.025 and the criteria is met
    • if l = £215K, then P(L>l) = 0.0006 ≤ 0.025 and the criteria is met
    • if l = £320K, then P(L>l) = 0 ≤ 0.025 and the criteria is met
    Of these bullets, the smallest loss that satisfies the criteria is l = £110K.

    ie VaR(97.5%) = £110K.

    We can interpret the result by saying there is a 97.5% chance that losses will be no greater than £110K.

    Re Tail Value at Risk, we can use a similar approach, ie start with the definition from Sweeting:

    TVaR(97.5%) = E[L | L > VaR(97.5%)]

    Portfolio A

    VaR(97.5%) = –£100K

    Losses, L, are:
    • £2m prob 0.02
    • –£100K prob 0.98
    We want E[L | L > –£100K]

    Lots of ways of doing this, I've done a sort of weighted average below:

    E[L | L > –£100K] = (£2m * 0.02 + –£100K * 0.005) / 0.025 = £1.58m

    Portfolio B

    VaR(97.5%) = £110K.

    Losses, L, are:
    • –£100K prob 0.6676
    • £5K prob 0.2725
    • £110K prob 0.0528
    • £215K prob 0.0065
    • £320K prob 0.0006
    • £X prob 0
    We want E[L | L > £110K]

    Lots of ways of doing this, I've done a sort of weighted average below:

    E[L | L > –£100K] = (£320K * 0.0006 + £215K * 0.0065 + £110K prob 0.0179) / 0.025 = £142.34K

    Hope this helps.
    Anna
     
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  6. Dar_Shan0209

    Dar_Shan0209 Ton up Member

    It does help a lot! Thanks Anna, happy to know that there are many ways that the examiner would have accepted but your method seems okayish for me.
     

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