Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 27J / a

Solution

After a record of value

v

, the first subsequent observation exceeding it has the original distribution conditioned on

X > v

. Indeed, summing over the number of failed observations gives the density

\sum_{j = 0}^{\infty} F (v)^{j} f (w) 1_{w > v} = \frac{f ( w )}{1 - F ( v )} 1_{w > v} .

Independence makes this conditional transition depend only on the latest record. Multiplying these transitions, starting with the density

f (v_{1})

, proves

f_{V_{1}, \dots, V_{n}} = 1_{0 < v_{1} < \dots < v_{n}} f (v_{1}) j = 2 \prod n \frac{f ( v _{j} )}{1 - F ( v _{j - 1} )} .

The ordering indicator is part of the density; it is zero elsewhere. Continuity of the observation distribution avoids ties.

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