I have the following stochastic process involving Brownian motions:
$Y(t)=W(t)-tW(1)$
Where $W(t)$ is a Brownian motion. Now if we consider:
$\mathbb{E}[Y(t_{1})(Y({t_2})-Y(t_{1}))]$
Can we rewrite this as:
$\mathbb{E}[Y(t_{1})(Y({t_2})-Y(t_{1}))]=\mathbb{E}[Y(t_{1})]\mathbb{E}[(Y{t_2})-Y(t_{1})]$
If not, how could we solve this expectation?
Lemma: $cov(Y_{u}, Y_{v}) = \min(u, v) - uv$. This is a covariance of Brownian brigde. Indeed, $cov(Y_{u}, Y_{v}) = cov(W(u)-uW(1), W(v)-vW(1))$ $=cov(W(u),W(v)) - u \cdot cov(W(1),W(v))-v \cdot cov(W(u), W(1))+uv \cdot cov(W(1),W(1)) = $ $=min(u,v) - u \min(1,v) - v \min(u,1)+uv\min(1,1) = \min(u, v) - uv.$
Suppose that $\mathbb{E}[Y(t_{1})(Y({t_2})-Y(t_{1}))]=\mathbb{E}[Y(t_{1})]\mathbb{E}[(Y_{t_2})-Y(t_{1})]$. Put $t_2 = 1$ and suppose that $0 < t_1 < t_2=1$. Then $Y(t_2) = 0$ and hence $Y(t_2)-Y(t_1) = -Y(t_1)$. Hence $\mathbb{E}[Y(t_1)(-Y(t_1))] = \mathbb{E}[Y(t_1)] \mathbb{E}[-Y(t_1)]$ thus $\mathbb{E}[Y^2(t_1)] = \mathbb{E}^2[Y(t_1)]$ and $\mathbb{D}[Y(t_1)]=0$. But $\mathbb{D}Y_{t_1} = cov (Y_{t_1}, Y_{t_1}) \ne 0$ because $cov(Y_{t_1}, Y_{t_1}) = \min(t_1, t_1) - t_1^2 \ne 0$ by lemma. We got a contradiction.
So we should use another way. $$\mathbb{E}[Y(t_{1})(Y({t_2})-Y(t_{1}))] =$$ $$= cov(Y_{t_1}, Y_{t_2}) - cov(Y_{t_1}, Y_{t_2}) = \min(t_1, t_2) - t_1t_2 - [\min(t_1, t_1) - t_1^2]$$ $$= t_1^2 - t_1 t_2 - t_1 + \min(t_1, t_2)$$