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VaR Question

Posted: March 29th, 2014, 9:49 pm
by Dantas
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

VaR Question

Posted: March 30th, 2014, 8:00 pm
by ppauper
if you look up Z=3.088, you get 0.001

VaR Question

Posted: March 30th, 2014, 11:11 pm
by Dantas
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?

VaR Question

Posted: March 31st, 2014, 7:57 am
by ACD
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.

VaR Question

Posted: March 31st, 2014, 8:36 am
by ppauper
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

VaR Question

Posted: March 31st, 2014, 9:11 am
by ppauper
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

VaR Question

Posted: April 1st, 2014, 12:07 am
by Dantas
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

VaR Question

Posted: April 1st, 2014, 7:35 am
by ppauper
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%