Jorion, Chapter 6 - Figure 6-3

MissJaguar

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Hello,

referring to the Study Notes Jorion Ch. 6 , page 5 & MR Book 2018 p. 51:

For model p=3%, when I calculate the (4-7.5)/ SQRT (0.03*0.97*250) ) I get = - 1.29 which is in range of cut off range for 97% two sided test, z value = + / - 2.17. Why is it then poor model?

Thanks
 

David Harper CFA FRM

David Harper CFA FRM
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@MissJaguar is your question not about any of the 7xx.x questions? you don't reference 712.1, eg. Sorry, it's just really time consuming to try and understand what's even being asked
@Nicole Seaman when the forum is upgraded (and the move function) fixed, can we move the above question to general T6. I finally figured out that it's a question about Jorion's chapter (although I don't have GARP's originals so the page number references don't help me). Time consuming ....
 
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David Harper CFA FRM

David Harper CFA FRM
Staff member
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@MissJaguar I figured out that you are asking about Jorion's Figure 6-3. Please note that your formula using the normal approximation (a separate debate when n*p = 7.5 < 10; conventionally, we prefer n*p > 10 and n*(1-p) > 10 before relying on the normal to approximate the binomial; Jorion doesn't mention this, but his example has n*p > 12.5 so he's fine) to test the null hypothesis that p = 3% given an observation of 4 exceptions. That's akin to "the null is that we have a good 97.0% VaR model," and observing 4.0 would not lead to a rejection, per your low test statistic, so the finding is "we cannot reject our null, so we decide that our 97.0% model is good!" But that's not exactly what Jorion is trying to illustrate with Fig 6-3. His Figure 6-3 means to illustrate a situation where we think we have a 99.0% VaR but we unwittingly have a 97.0% VaR model (aka, 97% is used to illustrate a "bad 99.0% model" but the dilemma is that there are many "bad 99.0% models" including 98%, 96%, 95% etc). So the cutoff of 4 is based on applying a decision rule in the context of assuming a 99.0% VaR model (i.e., assuming the expected number of exceptions is 1.0% * 250 = 2.5). Then this "mistaken cutoff" is used, in Fig 6-3, against the actual (true) distribution to show that, given the misunderstanding assumed, it is not greatly difficult to make a Type 2 error (to accept the null when the null is not true). Thanks,
 

MissJaguar

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David, you are awesome and should be on GARP Board!!

I think the list on page 7 of your materials is someone incorrect as I though cdf (i) = cdf (i-1) + pmf (i) and for e.g. cdf of 19 exceptions does not correspond cdf of 18 and pmf of 19
 
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