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Let $B$ be an brownian motion and let $s \leq t$. Compute $\mathrm { E } \left[ B _ { s } B _ { t } ^ { 2 } \right]$.

I know, that the answer is $0$, but I can't see how this ends being $0$.

My attempt: $\mathrm { E } \left[ B _ { s } B _ { t } ^ { 2 } \right] = \mathrm { E } \left[ B _ { s } \left( B _ { t } - B _ { s } + B _ { s } \right) ^ { 2 } \right] = \mathrm { E } \left[ B _ { s } \left( B _ { t } - B _ { s } \right) ^ { 2 } \right] + 2 \mathrm { E } \left[ B _ { s } ^ { 2 } \left( B _ { t } - B _ { s } \right) \right] + \mathrm { E } \left[ B _ { s } ^ { 3 } \right].$

So, my question is: Because of independence, can I split this up?:

$$\mathrm { E } \left[ B _ { s } \left( B _ { t } - B _ { s } \right) ^ { 2 } \right] = \mathrm { E } \left[ B _ { s } \right] \cdot \mathrm { E } \left[ \left( B _ { t } - B _ { s } \right) ^ { 2 } \right].$$ Then I can use some of the basic Brownian motion proberties. If $\mathrm { E } \left[ B _ { s } \right] = 0$, then the whole first term is zero. My first thought was that $\mathrm { E } \left[ B _ { s } \right] = 0$, but now I'm not sure why this is true.

I can do the similar things with second term.

The third term is zero because of the rule: $\mathrm { E } \left[ X_t^{2k+1} \right] = 0$

Any help is appreciated.

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  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Sep 4 '18 at 13:03
  • $\begingroup$ Use first that the "increments" of the Brownian motion associated to disjoined intervals (and limits of them) are independent. What can be said about the first two terms? The third term (in the attempt, last expression,) depends only on $B_s$, and we know the density... $\endgroup$ – dan_fulea Sep 4 '18 at 13:07
  • $\begingroup$ i have added more content now. thanks for feedback. $\endgroup$ – gariban17 Sep 4 '18 at 13:19
  • $\begingroup$ It looks like you have solved the whole problem already! Your explanations for why all three terms are zero are correct. $\endgroup$ – Mike Earnest Sep 4 '18 at 13:29
  • $\begingroup$ Thanks @MikeEarnst. But I could'nt find any theroms about "splitting integrals" like this. Could you help me finding some litterature about this? $\endgroup$ – gariban17 Sep 4 '18 at 13:40
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What you did is correct! The expected value of both $B_s$ and $B_t-B_s$ are zero, because by the rules of brownian motion they are centered Gaussian random variables with variance respectively $s$ and $t-s$. Due to the fact that they are centered Gaussians, the expected values are null.

So as you said the third term vanishes because it is the third moment of a centered gaussian, which is zero.

The first and the second are splittable because of the fact that the Brownian motion has independent increments, so $B_s$ and $B_t-B_s$ are independent.

In the first case you have $\mathbb{E}[B_s(B_t-B_s)^2]=\mathbb{E}[B_s]\mathbb{E}[(B_t-B_s)^2]=0$ because the first factor is null.

In the second case you have $\mathbb{E}[B_s^2(B_t-B_s)]=\mathbb{E}[B_s^2]\mathbb{E}[(B_t-B_s)]=0$ because the second factor is null, as said before.

So the expected value is zero overall

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  • $\begingroup$ how can I know that $B_s$ and $B_t - B_s$ is independent. What is the general rule? thanks for your answer. it was very helpfull. $\endgroup$ – gariban17 Sep 5 '18 at 10:58
  • $\begingroup$ There are many (equivalent) ways to define a Brownian motion, and one reports directly the independence that you need, that is: A Brownian motion is a Stochastic Process $(B_t)_{t\in\mathbb{R}}$ such that: - $B_0=0.$ - For every $0=t_0 \leq t_1\leq \dots \leq t_n$ the vector $(B_{t_1}-B_{t_0},\dots,B_{t_n}-B_{t_{n-1}})$ is a Gaussian vector of independent random variables with variance $(t_1,\dots,t_n-t_{n-1}).$ - It has continuous trajectories. $\endgroup$ – Riccardo Ceccon Sep 5 '18 at 14:21
  • $\begingroup$ Applying the second characteristic with $t_1=s$ and $t_2=t$ you obtain the independence between $B_s$ and $B_t-B_s.$ :) For other information: en.wikipedia.org/wiki/… $\endgroup$ – Riccardo Ceccon Sep 5 '18 at 14:27
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The shortest proof is that the law of $B$ is preserved by flipping to $-B$, but that changes your product to its negative. So your variable has the same mean as its negative, so this mean must be zero. This does assume you know the product is integrable.

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