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I am having some difficult showing what the following equals, where $x$ and $y$, $x>y$, distinct times:

$\mathbb{E}[\Delta W_x \Delta W_y]$

where each $\Delta W_t = W_t - W_{t-1}$.

I have decomposed it into its four terms, which allows taking expectations of the product of some W term and another, which I think is 0 by independence, but that makes this entire expectation 0, which makes the overall covariance I am solving for to be 0, which seems off to me.

The full problem I am trying to solve is:

$Cov(\Delta Z_t + \Delta\epsilon_t, \Delta Z_{t-i} + \Delta\epsilon_{t-i})$,

where $Z_t = \kappa W_t$ and $i = 1,2,3,...$, $W$ and $\epsilon $ independent, so guidance with how to would be even more appreciated, really.

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1 Answer 1

up vote 3 down vote accepted

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$ \mathbb{E}[\Delta W_x \Delta W_y] = 0 $$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ \mathbb{E}[(W_x - W_y + W_y - W_{x-1}) (W_y - W_{x-1} + W_{x-1} -W_{y-1})] \\ =&\ \mathbb{E}[(W_y - W_{x-1})^2] = y-x+1 \end{align} where the second to last step uses the independent increments property and the last step uses the fact that the second moment of Brownian motion is equal to the time step.

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Very helpful--thanks! One quick question: can this result also be extended to $x=y$ since $\mathbb{E}[W_t^2]=t$ and $\mathbb{E}[W_t W_s]=s, \ s < t $, by getting: $x - (x-1) + (x-1) + (x-1) = 1$? Just was wondering if the upward boundary of y was non-inclusive of x because I originally said $x>y$ or because of some other reason. Thanks! –  Michael Clinton Apr 14 at 10:07
    
The upper bound was just because the question required $x > y$. The statement holds for x=y. Actually, it trivially holds by the definition of Brownian motion (Variance/second moment equal to time step). –  pbr142 Apr 14 at 11:08
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