Mathematics Asked by TTL on November 1, 2021
Let $(B_t)_{tge 0}$ be the standard Brownian motion.
I am thinking about "translation" in Wiener integral. For exemple we have, for a non random constant $c$,$s,tin mathbb{R}$, $$int_{s}^{t+s} ccdot dB_u=c(B(t+s)-B(s))overset{d}=cB(t)=int_0^tccdot dB_u$$ from stationarity of the increments.
What’s happen if we take, more generally a $L^2(mathbb{R})$ (non-random) function $f$ ?
Do we have that for any $fin L^2(mathbb{R}),(s,t)in mathbb{R}^2$, $$int_{s}^{t+s} f(u)dB_uoverset{d}=int_0^t f(u+s)B_u.$$
Obviously the method is to prove it first for step functions of the form $f=sum a_i bf{1}_{(alpha_i,alpha_{i+1})}$ but I have to be careful because I would have a sum, so I cannot do everything I want.
I suppose that the result is true.
Well the easiest way to see this is to note that $$ int_s^{t+s} f(u),dB_usim Nleft(0,int_{s}^{t+s}f(u)^2,duright)=Nleft(0,int_{0}^{t}f(u+s)^2,duright)sim int_0^{t} f(u+s),dB_u, $$ provided that $f$ is deterministic. If $f$ is merely predictable, then you need to go back to the Riemann sums.
Answered by julian on November 1, 2021
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