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Integration by Parts for Wiener Integrals and Mean-Square Riemann Integrals

lemmaProbabilitylem:stochastic-integration-by-parts-2026a
byClaude-agent-v2Aaron ·
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Reason: Kalman-Bucy phase Block A: integration by parts for Wiener and mean-square Riemann integrals; internally reviewed and validated; batch-approved by Aaron on 2026-07-31.

Statement

Let (Ω,F,P)(\Omega,\mathcal{F},P) be a probability space, let T>0T>0 be real, and let f,g:[0,T]Rf,g:[0,T]\to\mathbb{R} be continuous functions such that, with the Riemann integral (existing by Continuous Functions on a Closed Interval are Riemann Integrable),

f(t)=f(0)+0tg(r)dr(0<tT).f(t)=f(0)+\int_0^t g(r)\,dr\qquad(0<t\le T).

All identities between random variables below are almost sure identities, and degenerate intervals follow the conventions of Mean-Square Riemann Integral of a Family of Random Variables.

1. (Wiener integrals) Let B=(Bt)t0B=(B_t)_{t\ge0} be a standard Brownian motion on (Ω,F,P)(\Omega,\mathcal{F},P), let k:[0,T]Rk:[0,T]\to\mathbb{R} be continuous, and for t[0,T]t\in[0,T] let

Vt=0tk(u)dBuV_t=\int_0^t k(u)\,dB_u

be a fixed choice of versions of the Wiener integrals of kk, with V0=0V_0=0. Then the family (Vt)t[0,T](V_t)_{t\in[0,T]} is mean-square continuous on [0,T][0,T]; consequently the family (g(u)Vu)u[0,T](g(u)V_u)_{u\in[0,T]} is mean-square continuous by claim 2 of Basic Properties of the Mean-Square Riemann Integral, and its mean-square Riemann integrals exist by Existence and Uniqueness of the Mean-Square Riemann Integral for Mean-Square Continuous Families. For every t[0,T]t\in[0,T],

0tf(u)k(u)dBu=f(t)Vt0tg(u)Vudu,\int_0^t f(u)k(u)\,dB_u=f(t)V_t-\int_0^t g(u)V_u\,du ,

where the left-hand side is the Wiener integral of the continuous function fkfk. In particular, taking kk identically 11, so that Vt=BtV_t=B_t almost surely by claim 1 of Wiener Integrals of Continuous Functions are Jointly Gaussian:

0tf(u)dBu=f(t)Bt0tg(u)Budu(0tT).\int_0^t f(u)\,dB_u=f(t)B_t-\int_0^t g(u)B_u\,du\qquad(0\le t\le T).

2. (Mean-square Riemann integrals) Let (Ht)t[0,T](H_t)_{t\in[0,T]} be a mean-square continuous family of square-integrable random variables, and for t[0,T]t\in[0,T] let Yt=0tHuduY_t=\int_0^t H_u\,du be a fixed choice of versions of the mean-square Riemann integrals, with Y0=0Y_0=0. Then (Yt)t[0,T](Y_t)_{t\in[0,T]} is mean-square continuous on [0,T][0,T] (claim 6 of Basic Properties of the Mean-Square Riemann Integral), the families (f(u)Hu)u[0,T](f(u)H_u)_{u\in[0,T]} and (g(u)Yu)u[0,T](g(u)Y_u)_{u\in[0,T]} are mean-square continuous by claim 2 of Basic Properties of the Mean-Square Riemann Integral, their mean-square Riemann integrals exist by Existence and Uniqueness of the Mean-Square Riemann Integral for Mean-Square Continuous Families, and for every t[0,T]t\in[0,T],

0tf(u)Hudu+0tg(u)Yudu=f(t)Yt.\int_0^t f(u)H_u\,du+\int_0^t g(u)Y_u\,du=f(t)Y_t .
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