Intuition on Expectation of Brownian motion starting at $x \in \mathbb{R}$ at a time $\epsilon>0$

Question

If we consider a Brownian motion starting at large $$ say $x=10^8$ i.e $B_0=x \text{ -}P^{x}$ almost surely then why $E^x[B_{\epsilon}]=0$ for $\epsilon$ very small. Wouldn't the expectation be close to $x$. Is it zero beacuse the measure $P^x$ assigns lower weights to values $B_{\epsilon}(\omega)$ near $x$.

It seems a bit Bm has continuous paths.

I apologize if the question is not well formulated.

score 1 · Accepted Answer · answered Apr 23 '18 at 18:54

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If your Brownian motion starts from $x$, then at time $\varepsilon > 0$ we have that $B_\varepsilon \sim N(x, \varepsilon)$. In particular $0 \neq \mathbb{E}^x[B_\varepsilon] = x$.

answered Apr 23 '18 at 18:54

Rhys Steele

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Thank you. I should have read the construction of the Brownian motion starting at $x$ more carefully from Karatzas and shreve – user3503589 Apr 23 '18 at 19:36

Intuition on Expectation of Brownian motion starting at $x \in \mathbb{R}$ at a time $\epsilon>0$

1 Answers1