Gauge Integral Structures for Stochastic Calculus and Quantum Electrodynamics. Patrick Muldowney

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Gauge Integral Structures for Stochastic Calculus and Quantum Electrodynamics - Patrick Muldowney

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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 alpha Subscript d Baseline equals 1 minus alpha Subscript u. Taking x left-parenthesis 0 right-parenthesis equals 10, let upper B element-of script upper A be the set consisting of the 16 elements

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 comma x left-parenthesis s right-parenthesis equals x left-parenthesis s minus 1 right-parenthesis plus-or-minus 1 for s equals 1 comma 2 comma 3 comma 4 semicolon

      and let upper P be an atomic measure with upper P left-parenthesis StartSet y EndSet right-parenthesis equals 0 if y not-an-element-of upper B, and, for x element-of upper B,

      (2.16)upper P left-parenthesis StartSet x EndSet right-parenthesis equals beta 1 beta 2 beta 3 beta 4 where beta Subscript s Baseline equals StartLayout Enlarged left-brace 1st Row 1st Column alpha Subscript u Baseline if x left-parenthesis s right-parenthesis 2nd Column equals 3rd Column x left-parenthesis s minus 1 right-parenthesis plus 1 comma 2nd Row 1st Column alpha Subscript d Baseline if x left-parenthesis s right-parenthesis 2nd Column equals 3rd Column x left-parenthesis s minus 1 right-parenthesis minus 1 EndLayout

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