Let B(·) and W (·) are two independent Brownian motions. Show two integrals have the same distributions.

Question

Let B(·) and W (·) are two independent Brownian motions. How to show that the distributions of $\int_{0}^{1}(B(t)+W(1-t))^2dt$ and $\int_{0}^{1}((B(t))^2+(B(1)-B(t))^2)dt$ are the same? I think that it suffices to show their characteristic functions are the same, but I don't know how to compute them. Can anyone help me with this?

Answer 1

Are these distributions the same? Consider for example the discretization of the integrals based on the subdivision $(0,\frac12,1)$. The associated Riemann sums are proportional to $$ 2X^2+Y^2+2Z^2+T^2+2XY+2XZ+2ZT\quad\mbox{and}\quad 3X^2+4XY+3Y^2, $$ respectively, where $(X,Y,Z,T)$ is i.i.d. standard normal. The ranks of these quadratic forms are $3$ and $2$ respectively, hence their distributions do not coincide. Of course, this discrepancy might disappear in the limit of subdivisions of smaller and smaller mesh, but does it?