What's new

# P.2 T5 Dowd study notes pg 10 confidence intervals

#### Stella G

##### New Member
Hello David,

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]?

Many thanks.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
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 ....

.... he shows the calculation. Although mine is generated in its own excel (see sheet here https://www.dropbox.com/s/cpxnh4ituxnvpd9/0204-dowd-3-27.xlsx?dl=0), as below. For Bin Width = 0.10 (ie, first column):
• f(q) = "Probability within bin" = 1 - 94.463% - 4.505% = 1.032%
• Per formula 3.27, Lower CI = 1.645 - 1.645*sqrt[4.505%*(1-4.505%)/1000]/1.032% = 0.600, and
• Upper CI = 1.645 + 1.645*sqrt[4.505%*(1-4.505%)/1000]/1.032% = 2.690. I hope that's helpful!

#### anniedu

##### New Member
Why divided by f(q) the probability within the bin?

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @anniedu Dowd shows the derivation (see below) of the standard error of the quantile estimator on page 70:

Thank you!

#### txiong

##### New Member
Hi @David Harper CFA FRM

Thank you for the example, I was wondering why p is equal to 4.505% in the example for the calculation of confidence interval.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
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,

#### txiong

##### New Member
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.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
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,

### Similar threads

Replies
1
Views
366
FAQ Before Exam FRM Part 2 Study Strategies
Replies
21
Views
30K