Gauge Integral Structures for Stochastic Calculus and Quantum Electrodynamics. Patrick Muldowney. Читать онлайн. Newlib. NEWLIB.NET

Автор: Patrick Muldowney
Издательство: John Wiley & Sons Limited
Серия:
Жанр произведения: Математика
Год издания: 0
isbn: 9781119595526
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alt="2 Superscript 16"/> members, one of which (for example) is

upper A equals StartSet left-parenthesis upper D comma upper U comma upper U comma upper D right-parenthesis comma left-parenthesis upper U comma upper D comma upper D comma upper U right-parenthesis comma left-parenthesis upper D comma upper U comma upper D comma upper U right-parenthesis comma left-parenthesis upper U comma upper U comma upper U comma upper U right-parenthesis comma left-parenthesis upper D comma upper D comma upper D comma upper D right-parenthesis EndSet comma upper P left-parenthesis StartSet omega EndSet right-parenthesis equals one sixteenth

      for each omega element-of normal upper Omega. For upper A above, upper P left-parenthesis upper A right-parenthesis equals five sixteenths period

      To relate this probability structure to the shareholding example, let bold upper R Superscript 4 Baseline equals bold upper R times bold upper R times bold upper R times bold upper R, and let

      (2.14)f colon normal upper Omega right-arrow from bar bold upper R Superscript 4 Baseline comma f left-parenthesis omega right-parenthesis equals left-parenthesis left-parenthesis x left-parenthesis 1 right-parenthesis comma x left-parenthesis 2 right-parenthesis comma x left-parenthesis 3 right-parenthesis comma x left-parenthesis 4 right-parenthesis right-parenthesis comma

      using Table 2.4; so, for instance,

f left-parenthesis omega right-parenthesis equals f left-parenthesis left-parenthesis upper U comma upper D comma upper D comma upper U right-parenthesis right-parenthesis equals left-parenthesis 11 comma 10 comma 9 comma 10 right-parenthesis equals left-parenthesis x left-parenthesis 1 right-parenthesis comma x left-parenthesis 2 right-parenthesis comma x left-parenthesis 3 right-parenthesis comma x left-parenthesis 4 right-parenthesis right-parenthesis comma

      and so on. Next, let bold upper S denote the stochastic integrals of the preceding section, so for x equals left-parenthesis x left-parenthesis 1 right-parenthesis comma x left-parenthesis 2 right-parenthesis comma x left-parenthesis 3 right-parenthesis comma x left-parenthesis 4 right-parenthesis right-parenthesis element-of bold upper R Superscript 4,

bold upper S left-parenthesis x right-parenthesis equals integral Subscript 0 Superscript 4 Baseline z left-parenthesis s right-parenthesis d x left-parenthesis s right-parenthesis equals sigma-summation Underscript s equals 1 Overscript 4 Endscripts z left-parenthesis s minus 1 right-parenthesis left-parenthesis x left-parenthesis s right-parenthesis minus x left-parenthesis s minus 1 right-parenthesis right-parenthesis comma

      so bold upper S left-parenthesis x right-parenthesis gives the values w left-parenthesis 4 right-parenthesis of Table 2.4. As described in Section 2.3, the rationale for deducing the probabilities of outcomes bold upper S left-parenthesis x right-parenthesis, equals w left-parenthesis 4 right-parenthesis, from the probabilities on normal upper Omega is the relationship

upper P left-parenthesis w left-parenthesis 4 right-parenthesis right-parenthesis equals upper P left-parenthesis f Superscript negative 1 Baseline left-parenthesis bold upper S Superscript negative 1 Baseline left-parenthesis w left-parenthesis 4 right-parenthesis right-parenthesis right-parenthesis right-parenthesis period

      For this calculation to work in general, the functions involved (f and bold upper S) must be measurable. But that is no problem in this case since all the sets involved are finite. To illustrate the calculation, take w left-parenthesis 4 right-parenthesis equals negative 2. Then, referring to Table 2.4 whenever necessary,

StartLayout 1st Row 1st Column bold upper S Superscript negative 1 Baseline left-parenthesis w left-parenthesis 4 right-parenthesis right-parenthesis 2nd Column equals 3rd Column bold upper S Superscript negative 1 Baseline left-parenthesis negative 2 right-parenthesis 4th Column equals 5th Column StartSet left-parenthesis 9 comma 8 comma 7 comma 8 right-parenthesis comma left-parenthesis 11 comma 10 comma 11 comma 10 right-parenthesis EndSet comma 2nd Row 1st Column f Superscript negative 1 Baseline left-parenthesis bold upper S Superscript negative 1 Baseline left-parenthesis w left-parenthesis 4 right-parenthesis right-parenthesis right-parenthesis 2nd Column equals 3rd Column f Superscript negative 1 Baseline left-parenthesis bold upper S Superscript negative 1 Baseline left-parenthesis negative 2 right-parenthesis right-parenthesis 4th Column equals 5th Column StartSet left-parenthesis upper D comma upper D comma upper D comma upper U right-parenthesis comma left-parenthesis upper U comma upper D comma upper U comma upper D right-parenthesis EndSet comma 3rd Row 1st Column upper P left-parenthesis w left-parenthesis 4 right-parenthesis right-parenthesis 2nd Column equals 3rd Column upper P left-parenthesis f Superscript negative 1 Baseline left-parenthesis bold upper S Superscript negative 1 Baseline left-parenthesis negative 2 right-parenthesis right-parenthesis right-parenthesis 4th Column equals 5th Column two sixteenths period EndLayout

      As a further illustration, suppose we now take

      letting script upper A be the sigma‐algebra of Borel subsets of bold upper R Superscript 4. For each upper A element-of script upper A define upper P left-parenthesis upper A right-parenthesis as follows. Suppose 0 less-than alpha Subscript u Baseline less-than 1 and Скачать книгу