Hi can someone help me out?X and Y are normally distributed but subject to independent shocks:X= e+ne~ N(0;1)n= 0 with probability 0.991 -10 with probability 0.009Show that the 99% Value at Risk is 3.088 ?Suppose Y has the same distribution as X. Set an equal weight portfolio of X and Y, show the 1% portfolio VaR is 4.9.Thanks

QuoteOriginally posted by: ppauperif you look up Z=3.088, you get 0.001[/QBut why 0.001 and not 0.01?Can you point what am I missing?

Last edited by Dantas on March 30th, 2014, 10:00 pm, edited 1 time in total.

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.

QuoteOriginally posted by: DantasHi can someone help me out?X and Y are normally distributed but subject to independent shocks:X= e+ne~ N(0;1)n= 0 with probability 0.991 -10 with probability 0.009Show that the 99% Value at Risk is 3.088 ?Suppose Y has the same distribution as X. Set an equal weight portfolio of X and Y, show the 1% portfolio VaR is 4.9.Thanks>> n= 0 with probability 0.991>>-10 with probability 0.009there is a 0.009 probability that the distribution is normal with mean -10 and variance 1 and a 0.991 probability that the distribution is normal with mean 0 and variance 1 the losses are going to be massive from the part with mean -10 and variance 1you want the 0.99 out of 1.00 VaR which leaves 0.01 (1.00 less 0.99)0.009 of the 0.01 will come from the part with mean -10 and variance 1which leaves 0.001 will come from the part with mean 0 and variance 10.001/0.991 =0.001009081736 which is the 3.088 you have

with the 2 of them, if you assume everything is independent,there's a 0.000081=0.009^2 probability you lose -10 from the shocksa 0.017838=2*0.009*0.991 probability you lose -5 from the shocksa 0.9820981=0.991^2 probability you lose 0 from the shocksif you add these up, you will see that the 99% VaR will be at the 0.007919 (=0.99-0.9820981) out of 0.017838 value from the loss of 5 and 0.007919/0.017838= 0.4439399036so you're going to be adding a little bit back to the loss of -5 which gives you the 4.9

First of all than thanks a lot Ppauper and ACD. There is one thing that isn't clear to me yet. I thought that the VaR only concerned about the probability of attain a certain left tail threshold. The way i understood both of yours explanation is somehow the expected value of that risk portfolio, which in my mind is more likely a CVAR, meaning giving I'm at the left tail what is the expected size of my loss. Did it make sense ?Thanks once again

the 1% VaR is the threshold value such that the probability that the mark-to-market loss on the portfolio over the given time horizon exceeds this value is 1%

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