Expected value for Brownian motion E(B1|B2=1)

Asked Jan 18 '20 at 18:11

Active Jan 18 '20 at 19:46

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I am currently stuck on the following question: $\mathrm{E}\left[B_{1} | B_{2}=1\right]$. Is there any one who can assist me? I can't figure out:

$\mathrm{E}\left[B_{1} +B_{2} - B_{2}|B_{2}=1 \right]$.

Normally you can say that $B_{2}-B_{1}$ is independent on previous intervals and therefore can be written loose. But now there is $B_{1}-B_{2}$, anyone who can give a push in the right direction?

edited Jan 18 '20 at 19:46

asked Jan 18 '20 at 18:11

Frakso34

Determine E[B1|B2=1]. Please explain of derivation, so what is the expected value of B1, given that B2=1? – Frakso34 Jan 18 '20 at 18:19
Time inversion may help. – Brian Moehring Jan 18 '20 at 18:37
Not a formal proof... but doesnt it follow from symmetry that $E[B_1 \mid B_0 = 0, B_2 = 1] = \frac12$? – antkam Jan 18 '20 at 19:46
2

Possible way: $$\mathbb E[B_1\mid B_2]=\mathbb E[B_1\mid B_1+(B_2-B_1)]=\mathbb E[X\mid X+Y]=\frac{X+Y}{2}$$ where $X,Y$ are independent and identically distributed with finite mean. So $\mathbb E[B_1\mid B_2=1]=\frac12$. – NCh Jan 19 '20 at 08:41
Thank you for your explanation! – Frakso34 Jan 19 '20 at 08:56

Expected value for Brownian motion E(B1|B2=1)

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