# Prove the equality $\mathbb{E}[(X_s^2-\mathbb{E}[X_s^2])(X_t^2-\mathbb{E}[X_t^2])]=2\mathbb{E}[X_sX_t]^2$

Question: Assume $$X_t=e^{-at}\int_0^te^{au}dB_u,$$ where $B$ is a standard Brownian motion. Show that for any $t,s\in \mathbb{R}_+$, we have $$\mathbb{E}\big[(X_s^2-\mathbb{E}[X_s^2])(X_t^2-\mathbb{E}[X_t^2])\big]=2\mathbb{E}[X_sX_t]^2.$$ The proof shouldn't be calculating explicit out the Left hand side and Right hand side and find them equal.

My trail:

First I proved by Isometry that $X_t$ is Gaussian, i.e. $\sim\mathcal{N}(0,\frac{1}{2a}(1-e^{-2at}))$ by $$\mathbb{E}[X_t^2]=\mathbb{E}[(e^{-at}\int_0^te^{au}dB_u)^2]=e^{-2at}\int_0^te^{2au}du=\frac{1}{2a}(1-e^{-2at}).$$

Then $$\mathbb{E}\big[(X_s^2-\mathbb{E}[X_s^2])(X_t^2-\mathbb{E}[X_t^2])\big]=\mathbb{E}[X_s^2X_t^2]-\mathbb{E}[X_s^2]\mathbb{E}[X_t^2].$$ Then I don't know how to proceed.

1. Set $$M_t := \int_0^t e^{au} \, dB_u.$$ Show that the assertion is equivalent to $$\mathbb{E}\bigg[(M_s^2-\mathbb{E}(M_s^2))(M_t^2-\mathbb{E}(M_t^2))\bigg]= 2 (\mathbb{E}(M_s M_t))^2. \tag{1}$$
2. Show (or recall) that $(M_t)_{t \geq 0}$ and $(M_t^2-\mathbb{E}(M_t^2))_{t \geq 0}$ are martingales.
3. Since $M_s$ is centered Gaussian, we have $$\mathbb{E}(M_s^4) = 3 \mathbb{E}(M_s^2) \tag{2}$$ for all $s \geq 0$.
4. Fix $s \leq t$. Use the tower property of conditional expectation and step 2 to show that the left-hand side of $(1)$ equals $$\mathbb{E}\bigg[ (M_s^2-\mathbb{E}(M_s^2))^2 \bigg]$$ which in turn equals $$\mathbb{E}(M_s^4)- (\mathbb{E}(M_s^2))^2.$$
5. Use another time the tower property and step 2 to show that the right-hand side of $(1)$ equals $$2 (\mathbb{E}(M_s^2))^2$$ for all $s \leq t$.
6. Conclude from step 3-5 that $(1)$ holds.