Could you please let me know which are the relevant values you plug in the equation q+1.645*se(q) to obtain the 90% confidence intrerval [0.600, 2.690]?
Hi @Stella.gkotsi Good question, because our current note really does not explain this adequately, sorry (cc: @Nicole Seaman including another edit for this note). So it is implementing Dowd's 3.27 ....
Hi @txiong It's a function of the selected bin width, h, about which Dowd says (emphasis mine) "in addition, the quantile standard error depends on the probability density function f (.) – so the choice of density function can make a difference to our estimates – and also on the bin width h, which is essentially arbitrary." (Dowd MMR p 70).
My left column above recreates Dowd's Example 3.7 which assumes bin width h = 0.10 but leads to wide CIs. Says Dowd, "The confidence interval narrows if we take a wider bin width, so suppose that we now repeat the exercise using a bin width h = 0.2, which is probably as wide as we can reasonably go with these data. q now falls into the range 1.645 ± 0.2/2 = [1.545, 1.745]."
... so my right column reproduces that; eg., 1.645 + 0.2/2 = 1.745. Then =1-NORM.S.DIST(1.745,TRUE) = 4.051%. Just like in the first column, =1-NORM.S.DIST(1.695,TRUE) = 1 - N(1.695) = 4.505%, where 1.695 = N^(-1)(95.0%) + 0.10/2 where 0.10 is the bin width. I hope that's helpful,
Hi @David Harper CFA FRM , Thank you for the detail explanation. I should be more specific about my question. I’m sorry if this is a stupid question but I was wondering why we always use the upper part of the bin.
Hi @txiong We don't, I was merely illustrating with the upper bound, because the lower bound has the same margin of error. What is the point of this exercise? We have estimated a VaR, which is just a quantile of the distribution; in the case of the standard normal, N(0,1), the 95.0% VaR is 1.645 because 5.0% of the standard normal's area is greater than 1.645. However, we recognize the 1.645 is only an estimate, so this is an estimate to carve a confidence interval around the estimate! The confidence interval looks like:
1.645 +/- z(α)*se(q), where z(α) is the deviate based on confidence, and se(q) is the standard error.
In the right-column above, this CI is:
Lower bound: 1.645 - 1.645 * 0.301375 = 1.149
Upper bound: 1.645 + 1.645 * 0.301375 = 2.141
See how that is a 90.0% CI = {1.149, 2.141}, around 1.645. We could carve instead a 99.0% CI:
Lower bound: 1.645 - 2.58 * 0.301375 = 0.869
Upper bound: 1.645 + 2.58 * 0.301375 = 2.421, and this is a 99.0% CI= {0.869, 2.421}, around 1.645
but the se(q) of 0.301375 did not change. It is given by sqrt[p*(1-p)/n]/f(q) = sqrt[4.051%*(1-4.051%)/1,000]/[1-93.881% - 4.051%] = 0.301374972. Both upper/lower bins are used, as Dowd explains (emphasis mine) "The confidence interval narrows if we take a wider bin width, so suppose that we now repeat the exercise using a bin width h = 0.2, which is probably as wide as we can reasonably go with these data. q now falls into the range 1.645 ± 0.2/2 = [1.545, 1.745]. p, the probability of a loss exceeding 1.745, is 0.0405, and the probability of profit or a loss less than 1.545 is 0.9388." Thanks,
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