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Computation of the standard error of a coherent risk measure

Thread starter #1
Dear all,

studying the computation of se(q) for the confidence interval of a coherent risk measure (here VaR) in the GARP books, I noticed two inconsistencies.
1. f(q) is indicated as "= 1-0.9446-0.0450" while I believe it would only make sense to compute it as "f(q)=1-(0.9446-0.0450)", i.e. "f(q)=1-0.9446+0.0450" in order to get the probability to be in the tails of the distribution.

2. In the computation of se(q), p = 0.0450 for both the upper and the lower bounds of VaR. I believe it should be p=0.9446 for the lower bound of this distribution, since it refers to the probability of the lower bound of the interval (1.6) rather than to the probability of the upper bound of the interval (1.7). Taking twice the same p (for upper and lower bound) seems very counter-intuitive.

Any explanation for these two inconsistencies / questions?
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
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#2
Hi @Michael Mayer Since we don't have an XLS for this (Dowd's 3.5.1 Standard Errors of Quantile Estimators) , I just started one here at https://www.dropbox.com/s/6yulo5zsnibrfub/dowd-3-5-se-quantiles.xlsx?dl=0

Re your first question, his approach looks correct to me. He appears to be solving for f(q) which is the probability of 1.595 < z < 1.695; i.e., it's a small slice, not tails. So it could be given by =NORM.S.DIST(1.695, TRUE) - NORM.S.DIST(1.595, TRUE) = 0.010321; his 0.0104 is rounded due to the imprecision of 0.045. This is Pr(1.595 < z < 1.695) = Pr(z < 1.695) - Pr (z < 1.595) = 0.9549 - 0.9446. What he's showing is the equivalent because Pr(z < 1.695) = 1 - Pr(z > 1.695) = 1 - 0.0451, so Pr(1.595 < z < 1.695) = Pr(z < 1.695) - Pr (z < 1.595) = [1 - Pr(z > 1.695)] - Pr (z < 1.595) = 1 - 0.0451 - 0.9446. I don't quite understand your second point, sorry ....
 

SalinaMiao

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#3
In Dowd's example under evaluating estimators of risk measures in chapter 3, in the first bullet point, I don't understand how did he came up with "the probability of a loss exceeding 1.695 is 4.5%", and also "probability of a profit or loss less than 1.595 is 94.46%"?

Can someone care to explain? Thank you
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
#4
Hi @SalinaMiao I moved your question to this thread, see above. The assumptions to which you refer are simply the normal probabilities associated quantiles; just like Prob(Z > 1.645) = 5.0%, except Dowd is retrieving this for bin edges at 1.645 +/- (h/2). He defined bin width h = 0.10 (top of page 70), around 1.645, so we have 1.645 - 0.10/2 = 1.595... and 1.645 + 0.10/2 = 1.695. Then NORM.S.DIST(1.595,TRUE) = 94.46%; i.e. Pr(Z < 1.645 - 0.10/2) = 94.46%, and also, 1 - NORM.S.DIST(1.695, TRUE) = 1 - 95.49% = 0.0451%; i.e. Pr(Z > 1.645 + 0.10/2) = 0.0451%. I hope that clarifies,
 
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