Let $(B_t)_{t\in [0,1]}$ be the standard Brownian Motion. Define $\mathcal{G_t}$ as $(\mathcal{F}_t\vee \sigma (B_1))_+$. Prove that for all $0s\leq t\leq 1$, $$ \mathbb{E}(B_t-B_s|\mathcal{G}_s)=\frac{t-s}{1-s}(B_1-B_s).$$ I have no idea how to proceed.I would be very grateful if someone can give me some hints.

  • $\begingroup$ Is that subscript $+$ implie positive maximum? $\endgroup$
    – Chinny84
    Jun 9, 2015 at 6:44
  • $\begingroup$ That means the filtration is enlarged such that it becomes a right continuous filtration. $\endgroup$
    – Patissot
    Jun 10, 2015 at 9:13

1 Answer 1


You have to start proving the claim on the rationals $s=j/n$ and $t=k/n$. We we can so rewrite you expression as :
$$E[B_t-B_s|B_1-B_s]=E\left[\sum_{i=j}^{k-1}B_{(i+1)/n}-B_{i/n}{\large\mid}\sum_{i=j}^{n-1}B_{(i+1)/n}-B_{i/n}\right].$$ Noting $Y_i=B_{(i+1)/n}-B_{i/n}$, and $Y=\sum\limits_{i=j}^{n-1}B_{(i+1)/n}-B_{i/n}=B_1-B_s$, the above expression simplifies to :

$$E[B_t-B_s|B_1-B_s]=E\left[\sum_{i=j}^{k-1}Y_i{\large\mid}\sum_{i=j}^{n-1}Y_i\right]=E\left[\sum_{i=j}^{k-1}Y_i{\large\mid} Y\right]$$

Now notice that $E[\sum_{i=j}^{k-1}Y_i|Y]=\frac{k-j}{n-j}Y$ this is because, as $Y_i$ are i.i.d., we have (for all i):

$E[Y_i|Y]=g(Y)$ where $g$ is a unique (a.s.) measurable function of $Y$ by definition of conditional expectation.

To finish the work for the rationals it suffices to notice that $\frac{k-j}{n-j}=\frac{t-s}{1-s}$.

So the left task for you is to show that it works fine in the continuous time case.

Best regards

  • $\begingroup$ Thank for your help. I probably wouldn't find the idea alone. $\endgroup$
    – Patissot
    Jun 10, 2015 at 9:09

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