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In the following questions ${[B(t),t\geq 0]}$ is a standard Brownian motion process and $T_a$ denotes the time it takes to hit this process to hit $a$.

  1. What is the distribution of $B(s)+B(t)$, $s\leq t$

  2. How to compute $E[B(t_1)B(t_2)B(t_3)]$ for $t_1\leq t_2\leq t_3$

Answers:

  1. $B(s)+B(t)=2B(s)+ B(t)- B(s)$. Now $2B(s)$ is normal with mean 0 and variance $4s$ and $B(t)-B(s)$ is normal with mean 0 and variance $t-s.$ Because $B(s)$ and $B(t)-B(s)$ are independent, $B(s)+ B(t)$ is normal with mean 0 and variance $3s+t$.

  2. I know answer to second question is 0. But I don't know how it is calculated. If you know the answer, answer it.

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  1. Note that the variance of $2B(s)$ is $4s$; therefore the variance of $$B_s+B_t = 2B_s+(B_t-B_s)$$ equals $4s+(t-s)=t+3s$.
  2. For fixed $t_1 \leq t_2 \leq t_3$ set $$c := \mathbb{E}(B(t_1) B(t_2) B(t_3)).$$ Obviously, $c$ depends only on the distribution of the Brownian motion $(B_t)_{t \geq 0}$. Since $W_t := -B_t$ is also a Brownian motion, this means that $$c = \mathbb{E}(W(t_1) W(t_2) W(t_3))$$ which implies by the very definition of $W_t$ that $$c =\mathbb{E}(W(t_1) W(t_2) W(t_3))= - \mathbb{E}(B(t_1) B(t_2) B(t_3)) = -c.$$ Hence, $$c=0.$$
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  • $\begingroup$ I agree to your answer to question(1). $\endgroup$ – Dhamnekar Winod Jan 7 '17 at 14:02
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I am providing answer to question (2)

$E[B(t_1)B(t_2)(B_3)]=E[E[B(t_1)B(t_2)B(t_3)|B(t_1)B(t_2)]]$

$E[B(t_1)B(t_2)B(t_3)]=[E[B(t_1)B(t_2)E[B(t_3)|B(t_1)B(t_2)]]$

$E[B(t_1)B(t_2)B(t_3)]=E[B(t_1)B(t_2)B(t_2)]$

$E[(B(t_1)B(t_2)B(t_3)]=E[B(t_1)B^2(t_2)|B(t_1)]$

$E[B(t_1)B(t_2)B(t_3)]=E[B(t_1)E[B^2(t_2)|B(t_1)]]$

$E[B(t_1)B(t_2)B(t_3)]=E[B(t_1)[(t_2-t_1)+B^2(t_1)]]$...(*)

$E[B(t_1)B(t_2)B(t_3)]=E[B^3(t_1)+(t_2-t_1)E[B(t_1)]$

$E[B(t_1)B(t_2)B(t_3)]=0]$

Where the Equality (*) follows since given $B(t_1),B(t_2)$ is normal with mean $B(t_1)$ and variance $(t_2-t_1)$. Also $E[B^3(t_1)]=0$ since $B(t_1)$ is n0rmal witn mean $0$

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