Proving Zero Covariance of Brownian Motion

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 :

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

• What are your thoughts? – Math1000 Mar 29 '15 at 0:27
• @Math1000 This is a follow up question from my other question : math.stackexchange.com/questions/1209701/… . I know I can find the conditional distribution and go from there, but I want to know if we can prove it this way – dimebucker Mar 29 '15 at 3:01
• How is Cov$(X|Y)$ supposed to be defined? // Is it on purpose that the argument of Cov in your question is something like $A-B+B$? – Did Mar 29 '15 at 8:11
• @Did I am trying to solve for $E(W_t | W_u , W_s)$. The book I am reading proposes that this can be done using the increment trick where I should expand $W_t$ into $W_t - \frac{t-s}{u-s}(W_u-W_s)+\frac{t-s}{u-s}(W_u-W_s)$ So this is what I have done, but I get to the point where I have : $E( (W_t - W_s) - \frac{t-s}{u-s}(W_u-W_s)|W_s,W_u) + A$ where A is the correct answer according to the book. So I want to know how can i prove that I can drop the conditioning on $W_u,W_s$, and so the next logical step is to show that they are independent by showing that they have zero covariance – dimebucker Mar 29 '15 at 8:27
• @Math1000 Its a set of lecture notes i found online. Here is the link: disi.unal.edu.co/~gjhernandezp/mathcomm/slides/bm.pdf , the problem is on page 11/12 – dimebucker Mar 29 '15 at 12:43

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,

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

• What is it about joint gaussians that lets us say that independence implies joint independence? – dimebucker Mar 30 '15 at 4:33
• @dimebucker91, What property must the joint pdf of $F_{t,s}, W_s$ and $W_u$ satisfy to justify the independence you seek? Because we are working with a joint Gaussian pdf, do the computed covariations being $0$ imply that the joint pdf satisfies this property? – ki3i Mar 30 '15 at 11:44