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Let $\Delta B_j=B_{t_{j+1}}-B_{t_j}$ where $B_t$ is Brownian motion, and $e_i(\omega)$ measurable with respect to $\sigma(B_{t_i})$. In Oksendal's 'Stochastic Differential Equations' he states:

$$ E[e_ie_j\Delta B_i\Delta B_j]= \begin{cases} 0 & i\ne j \\ E[e_j^2] & i=j \end{cases} $$

Justifying this because '$e_ie_j\Delta B_i$ and $\Delta B_j$ are independent if $i<j$'. I am trying to understand this.

I understand that Brownian motion is normally distributed with independent increments, so: $$E[\Delta B_i\Delta B_j]=0$$ However, how can we justify $e_ie_j\Delta B_i$ being independent to $\Delta B_j$? Even if they are, how can we treat $E[e_ie_j\Delta B_i]$? Basically I believe my confusion stems from how to treat products of measurable functions.

Ruvi Lecamwasam
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2 Answers2

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For every $i$, let $F_i=\sigma(B_s,s\leqslant t_i)$ then each $e_i$ is $F_i$-measurable and each $\Delta B_i$ is centered, independent of $F_i$, and $F_{i+1}$-measurable.

Now, fix $i\lt j$. Then $e_i$, $e_j$ and $\Delta B_i$ are all $F_j$-measurable and $\Delta B_j$ is independent of $F_j$ hence $e_ie_j\Delta B_i$ and $\Delta B_j$ are independent, a fact which implies $$E(e_ie_j\Delta B_i\Delta B_j)=E(e_ie_j\Delta B_i)\cdot E(\Delta B_j)=E(e_ie_j\Delta B_i)\cdot0=0.$$ Thus, the argument does not require to know the value of $E(e_ie_j\Delta B_i)$.

Likewise, for every $i$, $e_i$ is $F_i$-measurable and $\Delta B_i$ is independent of $F_i$ hence $$E((e_i\Delta B_i)^2)=E(e_i^2)\cdot E((\Delta B_i)^2)=E(e_i^2)\cdot(t_{i+1}-t_i).$$

Did
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  • Hello Did. Sorry to ask you this so long after the question, but how exactly do we know that $\Delta B_j$ is independent of $\mathcal{F}j$?. Intuitively it makes sense, in that 'knowing' the value of $B_j$ does not give you any information about the increment $B{j+1}-B_j$, but I can't see how to show that mathematically. – Ruvi Lecamwasam Oct 30 '14 at 11:49
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    This is part of the very definition of Brownian motion. – Did Oct 30 '14 at 12:40
  • I have seen it defined that way, but in the book I am reading (Oksendal) he instead defines some probability measure, then uses the Kolmogorov extension theorem to state that there exists some corresponding stochastic process which we call 'Brownian Motion'.

    Using that Brownian motion is a Martingale, does it make sense to say that independence follows from: $E[\Delta B_j|\mathcal{F}j]=E[B{j+1}|\mathcal{F}_j]-E[B_j|\mathcal{F}_j]=E[B_j|\mathcal{F}_t]-E[B_j|\mathcal{F}_j]=0=E[\Delta B_j]$

     – Ruvi Lecamwasam Oct 31 '14 at 03:44
  • Being a martingale never implies the independence of increments and here you need the independence so no, the approach you suggest cannot work. If you insist on using O's presentation (which I haven't checked), I suggest checking the independence of increments at the step when one "defines some probability measure", as you put it, and then at every subsequent step. – Did Oct 31 '14 at 08:03
  • Ok, thank you very much for your help with that. – Ruvi Lecamwasam Oct 31 '14 at 22:12
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You can see it using the "tower property" for instance,

Assume $i<j$, then

$$E[e_i e_j \Delta B_i \Delta B_j] = E\left[ E[e_i e_j \Delta B_i \Delta B_j |\mathcal{F}_{t_j}]\right].$$

Now observe that since $i<j$ then $\mathcal{F}_{t_i}\subset \mathcal{F}_{t_j}$ and by hypothesis $e_i$ is $\mathcal{F}_{t_i}$-measurable and hence also $\mathcal{F}_{t_j}$-measurable, also $\Delta B_i = B_{t_{i+1}}-B_{t_i}$ is $\mathcal{F}_{t_j}$-measurable. Hence $$E[e_i e_j \Delta B_i \Delta B_j] = E\left[ e_i e_j \Delta B_i E[\Delta B_j |\mathcal{F}_{t_j}]\right].$$

Finally observe that $\Delta B_j$ is independent of $\mathcal{F}_{t_j}$ so $E[\Delta B_j |\mathcal{F}_{t_j}]=E[\Delta B_j]=0$ and hence

$$E[e_i e_j \Delta B_i \Delta B_j] = 0.$$

This was maybe very detailed but useful to understand why.

Martingalo
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