Proving Zero Covariance of Brownian Motion

Question

Let $u>t>s\ge0$

I Want to know whether the following statement:

$Cov[(W_t - W_s) -\frac{t-s}{u-s} (W_u-W_s),W_k] =0 ~~~~~~~~~~~~~~\forall K \in \{u,s\}$

implies :

$Cov[(W_t - W_s) -\frac{t-s}{u-s} (W_u-W_s),(W_u,W_s)] = 0$.

Answer 1

Define $F_{t,s}:=(W_t-W_s)-\frac{t-s}{u-s}(W_u-W_s)$. Your intention is to argue that the jointly Gaussian random variables $F_{t,s}, W_s$ and $W_u$ are such that $F_{t,s}$ is independent of $W_s, W_u$. Being Jointly Gaussian, it is enough to show that $\mathbb {C}ov(F_{t,s}, W_s) = \mathbb {C}ov(F_{t,s}, W_u) = 0$.

Well, $$ \mathbb {C}ov(F_{t,s}, W_s) = \mathbb E[F_{t,s}W_s]-\mathbb E[F_{t,s}]\mathbb E[W_s] = \Big(s-s-\frac{t-s}{u-s}(s-s)\Big) - 0\cdot0= 0 $$

and

$$ \mathbb {C}ov(F_{t,s}, W_u) = \mathbb E[F_{t,s}W_u]-\mathbb E[F_{t,s}]\mathbb E[W_u] = \Big(t-s-\frac{t-s}{u-s}(u-s)\Big) - 0\cdot0 = 0 $$

Therefore, $F_{s,t}$ is independent of $W_s, W_u$, from which we deduce $$\ \mathbb E[F_{t,s}\vert W_s, W_u] = \mathbb E[F_{t,s}] = 0\,.$$ That is,

$$ \mathbb E[W_t - W_s\vert W_s, W_u] = \frac{t-s}{u-s}\mathbb E [W_u - W_s\vert W_s, W_u] = \frac{t-s}{u-s}(W_u - W_s)\,.$$