I might not have a clear understanding for VaR yet. for question 8 in oct2011 exam. "investment strategy B involves inseting 100,000 in each of the 20 corporate bond issues fromt he portfolio. Schedule of CDF for the binomial (n=20,p=0.02) X F(x) 0 0.6676 1 0.9401 2 0.9929 3 0.9994 4 1.0000 Why is the 97.5% VaR determined at the 2 default level, rather than 1 default level? F(1) = 0.9401 and F(2) = 0.9929 Thanks.
I believe since the probability of one or fewer defaults is less than .975, at .975 2 defaults need to be assumed.
re At 2 defaults, the probability is 0.9902 which exceeds 0.975. If VaR is defined as the maximum defaults that can be exposed w/o exceeding 0.975, shouldn't 1 be used?
Lets say I run 10,000 simulations. For the 97.5% ci VaR, I want to order the 10,000 simulations from best (least defaults, lets call that simulation #1) to worse (most defaults - #10,000), then I take #9750. The 6676 best simulations had defaults = 0. The next 2800, upto #940, had 1 default. #9402 has two defaults, as does #9750, as does #9929. #9930 has three defaults, etc.
