I just bumped into the stochastic integral $$ \int_\rho^1 (W_t - W_{t-\rho}) dW_t $$ where $0 < \rho < 1$ is a constant and $W$ is a standard Wiener process. It would be nice if we have a closed-form representation of the cumulative distribution function, but for my work it suffices to show that there is some simple transform to make its law $\rho$-invariant, that is, to make its law not dependent on $\rho$.

Does it appear possible to any one? Thanks for any help in advance!

Additional Information: I guessed so because when $\rho$ is large enough, $\int_\rho^1 (W_t - W_{t-\rho})dW_t \approx W_\rho (W_1 - W_\rho)$. I tried to multiply it by $\frac{1}{\sqrt{\rho(1-\rho)}}$, but in some simple Monte Carlo experiments the distribution of the stochastic integral doesn't seem to be $\rho$-invariant.

I just bumped into the stochastic integral $$ \int_\rho^1 (W_t - W_{t-\rho}) \,dW_t $$ where $0 < \rho < 1$ is a constant and $W$ is a standard Wiener process. It would be nice if we have a closed-form representation of the cumulative distribution function, but for my work it suffices to show that there is some simple transform to make its law $\rho$-invariant, that is, to make its law not dependent on $\rho$.

Does it appear possible to any one? Thanks for any help in advance!

Additional Information: I guessed so because when $\rho$ is large enough, $\int_\rho^1 (W_t - W_{t-\rho})\,dW_t \approx W_\rho (W_1 - W_\rho)$. I tried to multiply it by $\frac{1}{\sqrt{\rho(1-\rho)}}$, but in some simple Monte Carlo experiments the distribution of the stochastic integral doesn't seem to be $\rho$-invariant.