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# Exercise Solution 14.8

1. Exceedence ti data is calculated from the data of Exhibit 14.13 and presented in Exhibit s1 below.
timetitimetitimetitimetitimeti
tat ttat ttat ttat ttat t
-1240-990-740-490-240
-1230-980-730-480-230
-1220-970-720-470-220
-1210-960-710-460-210
-1200-951-700-450-200
-1190-940-690-440-190
-1180-931-680-430-181
-1170-920-670-420-170
-1160-910-660-410-160
-1150-900-650-400-150
-1140-890-640-390-140
-1130-880-630-380-130
-1120-870-620-370-120
-1110-860-610-360-110
-1100-850-600-350-100
-1090-841-590-340-90
-1080-830-580-330-80
-1070-820-570-320-70
-1060-810-560-310-60
-1050-800-550-300-50
-1040-790-540-290-40
-1030-780-530-280-30
-1020-770-521-270-21
-1010-760-510-260-10
-1000-750-500-25000
Exhibit s1: Exceedence ti data calculated from the data of Exhibit 14.13. A 1 indicates an exceedence occurred. A 0 indicates one did not.

b.

1. To apply our recommended standard coverage test, we need
• the quantile of loss the value-at-risk measure is intended to measure: q = 0.99,
• the number of observations: α + 1 = 125, and
• the number of exceedences, from Exhibit s1 above: x = 6.

In Exhibit 14.3, we find that our recommended standard coverage test’s non-rejection interval for q = 0.99 and α + 1 = 125 is [0, 3]. Since our number of exceedances falls outside this interval, we reject the value-at-risk measure.

2. In Exhibit 14.4, we find that the PF test’s non-rejection interval for q = 0.99 and α + 1 = 125 is [0, 4]. Since our number of exceedances falls outside this interval, we reject the value-at-risk measure.
3. We cannot use the Basel Committee’s Trafic Light coverage test because the test applies only if there are α + 1 = 250 observations.