Let $(B_t)_{t\ge 0}$ be the standard Brownian motion. I am thinking about "translation" in Wiener integral. For exemple we have, for a non random constant $c$,$s,t\in \mathbb{R}$, $$\int_{s}^{t+s} c\cdot dB_u=c(B(t+s)-B(s))\overset{d}=cB(t)=\int_0^tc\cdot dB_u$$ 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 $f\in L^2(\mathbb{R}),(s,t)\in \mathbb{R}^2$, $$\int_{s}^{t+s} f(u)dB_u\overset{d}=\int_0^t f(u+s)B_u.$$
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 $$ \int_s^{t+s} f(u)\,dB_u\sim N\left(0,\int_{s}^{t+s}f(u)^2\,du\right)=N\left(0,\int_{0}^{t}f(u+s)^2\,du\right)\sim \int_0^{t} f(u+s)\,dB_u, $$ provided that $f$ is deterministic. If $f$ is merely predictable, then you need to go back to the Riemann sums.