Exercise Solution 14.8

Exceedence ^ti data is calculated from the data of Exhibit 14.13 and presented in Exhibit s1 below.

time	^ti	time	^ti	time	^ti	time	^ti	time	^ti
t	at t	t	at t	t	at t	t	at t	t	at t
-124	0	-99	0	-74	0	-49	0	-24	0
-123	0	-98	0	-73	0	-48	0	-23	0
-122	0	-97	0	-72	0	-47	0	-22	0
-121	0	-96	0	-71	0	-46	0	-21	0
-120	0	-95	1	-70	0	-45	0	-20	0
-119	0	-94	0	-69	0	-44	0	-19	0
-118	0	-93	1	-68	0	-43	0	-18	1
-117	0	-92	0	-67	0	-42	0	-17	0
-116	0	-91	0	-66	0	-41	0	-16	0
-115	0	-90	0	-65	0	-40	0	-15	0
-114	0	-89	0	-64	0	-39	0	-14	0
-113	0	-88	0	-63	0	-38	0	-13	0
-112	0	-87	0	-62	0	-37	0	-12	0
-111	0	-86	0	-61	0	-36	0	-11	0
-110	0	-85	0	-60	0	-35	0	-10	0
-109	0	-84	1	-59	0	-34	0	-9	0
-108	0	-83	0	-58	0	-33	0	-8	0
-107	0	-82	0	-57	0	-32	0	-7	0
-106	0	-81	0	-56	0	-31	0	-6	0
-105	0	-80	0	-55	0	-30	0	-5	0
-104	0	-79	0	-54	0	-29	0	-4	0
-103	0	-78	0	-53	0	-28	0	-3	0
-102	0	-77	0	-52	1	-27	0	-2	1
-101	0	-76	0	-51	0	-26	0	-1	0
-100	0	-75	0	-50	0	-25	0	0	0

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.

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.
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.
We cannot use the Basel Committee’s Trafic Light coverage test because the test applies only if there are α + 1 = 250 observations.