Exercice. Let $(B_t)_{t \in \mathbb{R}_{+}}$ be an brownien motion defined on a probability space $(\Omega, \mathcal{F}, \mathbb{R})$ with values in $\mathbb{R}$. Let $t > 0$. Prove that $\int_{0}^{t} B_{s}^{2} ~\mathrm{d}s$ and $t^2\int_{0}^{1} B_{s}^{2} ~\mathrm{d}s$ have the same law.
My attempt : I know that by change of variables $s \mapsto ts$, we have $\int_{0}^{t} B_{s}^{2} ~\mathrm{d}s = t^2 \int_{0}^{1} (\frac{1}{\sqrt{t}} B_{ts})^{2} ~\mathrm{d}s$. And I also know that $(\frac{1}{\sqrt{t}} B_{ts})_{s \in \mathbb{R}_{+}}$ is a centered gaussian process with covariance function $\mathbb{E}[\frac{1}{\sqrt{t}} B_{ts}\frac{1}{\sqrt{t}} B_{ts^{\prime}}] = s \wedge s^{\prime}$ and continuous sample paths, so it's a brownien motion.
How can I, from these, conclude that $\int_{0}^{1} B_{s}^{2} ~\mathrm{d}s$ and $\int_{0}^{1} (\frac{1}{\sqrt{t}} B_{ts})^{2} ~\mathrm{d}s$ have the same law ? Or is there a different proof ? Thanks.
If $(X_s)_{s\geqslant 0}$ and $(Y_s)_{s\geqslant 0}$ have the same distribution, then $\left(\int_0^t X_sds\right)_{t\geqslant 0}$ has the same distribution as $\left(\int_0^tY_sds\right)_{t\geqslant 0}$. To see this, one can show that the Riemann sums have the same finite dimensional distributions and the Riemann sums converge in distribution to the integrals.