VaR Backtest (2010 L1. Q39 modified)

[mgarcia]

On page 7 from the Oct 2nd, 2010 FRM Level 2 Webinar, it says:
•9 exceptions: p = ~6.7% (3.6% ~N) ...Fail to reject @ 95% 2008 & 2009: Accept VaR model as good

Looking at the table provided, the number 9 has a CDF of 96.9%, which is higher than the 95% confidence level.
I am confused, why did we failed to reject? Why 9:exceptions=~6.7% and not 3.1%?
Thanks

 

[David Harper CFA FRM]

Hi MGT,

Your idea is right, it's just that the discrete CDF is a less than or equal to; i.e.,
BINOM.DIST(X <= 8, 500 trials, p =1%, if CDF true) = 93.3%, and
BINOM.DIST(X <= 9, 500 trials, p =1%, if CDF true) = 96.9%

In this case, 93.3% is the probability of 8 or fewer exceptions, such that
P[9 or more] = 1 - P[8 or fewer], not P[9 or fewer]
... see how it's an issue only for the discrete CDF variable?

In this case, similarly, 10 exceptions does have a p value of 3.1% because 3.1% is the sum of each individual probability, sum of: P[X =10] + P[ X = 11] + .... = 1 - P[X <= 9]

David