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There's an equation in my script, which I do not understand. Let $(B_t)$ be a Brownian Motion and $\Gamma\in\mathcal{B}(\mathbb{R}^n)$, $t\ge s$ the equation is

$$P(B_t\in\Gamma | B_s)=\frac{1}{\sqrt{(2\pi)^d(t-s)^d}}\int_\Gamma \exp{\left(-\frac{|z-B_s|^2}{2(t-s)}\right)}dz$$

A reference why this is true would be appreciated. I think there is a Theorem which tells us how to calculate such an expression. Just for completeness, the only result I know is:

Let $\mathcal{G}\subset\mathcal{F}$ a $\sigma$-field, $X$ $\mathcal{G}$ measurable and $Y$ independent of $\mathcal{G}$. For every $F:S_1\times S_2\to[0,\infty]$ which is $\mathcal{S}_1\times\mathcal{S}_2$ measurable we have $$E[F(X,Y)|\mathcal{G}]=E[F(x,Y)]|_{x=X(\omega )}=g(X(\omega ))$$ where $g(x):=E[F(x,Y)]$.

However I do not see how to apply this in this situation.

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1 Answer 1

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Let $X=B_s$, $Y=B_t-B_s$ and $f(x,y):=(x+y)\chi_A$. All the conditions are satisfied.

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Why is in the exp function $|z-B_s|$ and not $|z+B_s|$ ? –  user20869 Jul 25 '12 at 12:23

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