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On a classic probability space with $B$ a brownian motion. How to prove for : $$0\leq s<t\leq 1,\ \ \ \mathbb E (B_t-B_s\mid B_1-B_s)=\frac{t-s}{1-s}(B_1-B_s)$$


What I have done: I am sure that we have the brownian filtration : $\mathcal F_s$ independant to $\sigma (B_1-B_s)$ and also $\mathcal F_t\vee \sigma (B_1) = \mathcal F_t\vee \sigma (B_1-B_s)$. So then we can write : $$\mathbb E (B_t-B_s\mid B_1-B_s)=\mathbb E (B_t-B_s\mid \mathcal F_s\vee \sigma (B_1))$$

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Define $X=B_t-B_s$ and $Y=B_1-B_s$. Now since $(X,Y)$ are jointly normaly distributed, with mean $0$ and variance $\Sigma$ you have $$E(X|Y=y)=E(X)+\frac{Cov(X,Y)}{Var(Y)}(y-E(Y))$$ which shall translate into your desired results.

Note $Cov(X,Y)=Cos(B_t-B_s,B_1-B_s)=t-s$ since $t\leq1$, and $Var(Y)=1-s$.

For joint normality you may look back at the properties of the Brownian Motion. For that see here. For conditional expectations see here.

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  • $\begingroup$ Do you have a proof of that (even an internet link) ? And how do you assert that $(X,Y)$ are jointly normaly distributed (they are not independant) ? $\endgroup$
    – Al Bundy
    Nov 14, 2016 at 8:41
  • $\begingroup$ A proof for joint normality or the conditional expectation? $\endgroup$
    – Math-fun
    Nov 14, 2016 at 8:42
  • $\begingroup$ Yes both please $\endgroup$
    – Al Bundy
    Nov 14, 2016 at 8:43
  • $\begingroup$ For $f$ a borelian function we can write : $B_1-B_s=f(B_1-B_t, B_t-B_s)$. Is it independant to $B_t-B_s$ ? I don't think so. $\endgroup$
    – Al Bundy
    Nov 14, 2016 at 8:49
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    $\begingroup$ Note that this is a property of a Brownian motion to have increments jointly normally distributed. $\endgroup$
    – Math-fun
    Nov 14, 2016 at 14:14

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