March 31st, 2014, 7:57 am
I don't think ppauper's answer is in the right area, X is not normally distributed in this problem it is a mixture of 2 normals. The inverse of the VaR function is:f(x) = 0.991 * cum.norm.dist(x,mean=0,std=1) + 0.001 * cum.norm.dist(x,mean=-10,std=1) So do a root find on:g(x) = 0.991 * cum.norm.dist(x,mean=0,std=1) + 0.001 * cum.norm.dist(x,mean=-10,std=1) -0.01You'll get a value of -3.088 to 3 d.p. which is your VaR taking the positive value of the result for convention. The second result presumably follows from there considering all of the possible mixtures.
Last edited by
ACD on March 30th, 2014, 10:00 pm, edited 1 time in total.