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The issue is to compute the conditional covariance of dependent Brownian motion increments:

$$\operatorname{Cov} \left(W(T)-W(t_i), W(T)-W(t_j)|\frac{1}{\alpha}\sum_{k=0}^{N-1}W(T)-W(t_k)\right),$$

where $\alpha$ is a normalizing constant, such that $Z=\frac{1}{\alpha}\sum_{k=0}^{N-1}W(T)-W(t_k)$ has a standard normal distribution $N(0,1)$,

$i,j \in \{0,...N-1\}, T=t_{N-1}, t_i=\frac{T}{N}i, t_j=\frac{T}{N}j$

and $W=\{W(t)\}_{t\in [t_0,t_{N-1}]}$ is a standard Brownian motion.

Since the Brownian motion increments $W(T)-W(t_i)$ and $W(T)-W(t_j)$ are not mutually independent, this question seems not trivial to me. I would appreciate any help.

Edit #1: terms, which I think, I have managed to compute

Let $X=W(T)-W(t_i)\sim N(0, T-t_i)$, $Y=W(T)-W(t_j)\sim N(0, T-t_j)$ and $Z=\frac{1}{\alpha}\sum_{k=0}^{N-1}W(T)-W(t_k)\sim N(0,1)$. Then, we have to compute $$Cov(X,Y|Z)=E[XY|Z]-E[X|Z]E[Y|Z].$$ Next, I claim that $X,Z$ and $Y,Z$ are bivariate normal. But I do not know, how to show this. Any help is appreciated. If it is true, then $$E[X|Z]=E[X]+\frac{Cov(X,Z)(Z-E[Z])}{Var[Z]}=Cov(X,Z)Z,$$ since $E[X]=E[Y]=E[Z]=0$, $Var[Z]=1$. Now, $$Cov(X,Z)=E[XZ]=\frac{1}{\alpha}(T-t_i + \sum^{N-1}_{k\neq i} T-t_i-t_k+t_k\wedge t_i).$$ Analogously, we obtain $E[Y|Z]$. Next, we need to compute $$E[XY|Z] = E[XY] + Cov(XY,Z),$$ which is true if $XY$ and $Z$ are bivariate normal. Is it correct, that $XY$ and $Z$ are bivariate normal?

Then, given the statement above is correct, we find that $$E[XY]=T-t_i-t_j+t_i\wedge t_j$$ and $$Cov(XY,Z)=E[XYZ]=0$$ as the third moment of a standard normal distribution. Is this also correct?

At the end, under all assumptions I have made, we obtain $$Cov(X,Y|Z)=E[XY] - Cov(X,Z)ZCov(Y,Z)Z=T-t_i-t_j+t_i\wedge t_j - \frac{Z^2}{\alpha^2}(T-t_i + \sum^{N-1}_{k\neq i} T-t_i-t_k+t_k\wedge t_i)(T-t_j + \sum^{N-1}_{k\neq j} T-t_j-t_k+t_k\wedge t_j),$$ hence $Cov(X,Y|Z)$ is a random variable, since it depends on $Z^2$?

I would very appreciate correction of possible mistakes, complement and suggestion on the improvement of my results. Thank you very much beforehand!

Edit #2: Based on comments of Did.

Let $X=aZ + bU$, $Y=cZ+dU+eV$, where $U,V\sim N(0,1)$ and $U,V$ independent of $Z$.

Then, we know $Cov(U,Z)=E[UZ]=0$ and $Cov(V,Z)=E[VZ]=0$.

Hence, given $U=\frac{1}{b}(X-aZ)$ we get $$E[\frac{1}{b}(X-aZ)Z]=\frac{1}{b}E[XZ]-\frac{a}{b}=0 \implies a=E[XZ].$$ Using the fact that $Var[U]=E[U^2]=1$ we obtain $$E[\frac{1}{b^2}(X^2 - 2aXZ + a^2Z^2]\implies b=\pm\sqrt{T-t_i+a^2}.$$ Next, $V=\frac{1}{e}(Y-cZ-dU)$ and $Cov(V,Z)=E[VZ]=0$. Therefore, $$E[\frac{1}{e}(Y-cZ-dU)Z]=E[YZ]-c=0\implies c=E[YZ].$$ Moreover, $Var[V]=E[V^2]=1$. It follows that $$E[\frac{1}{e^2}(Y-cZ-dU)^2]=\frac{1}{e^2}(T-t_j+c^2+d^2-2c^2-2dE[YU]).$$ Let's compute $E[YU] = E[Y(\frac{1}{b}X-\frac{a}{b}Z)]=\frac{1}{b}E[XY]-\frac{ac}{b}.$ Then, we get $$T-t_j+c^2+d^2-2c^2-2\frac{d}{b}E[XY]+\frac{2acd}{b}=e^2.$$ At this point, can I say that, for example, $d=E[XY]$ and then, $$e=\pm\sqrt{T-t_j+d^2-c^2-2\frac{d^2}{b}+\frac{2acd}{b}}.$$ Finally, $$Cov(X,Y|Z)=bd=\sqrt{T-t_i+E[XZ]^2}E[XY]$$ for $i<j$ and $$Cov(X,Y|Z)= T-t_i+E[XZ]^2$$ for $i=j$.

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  • $\begingroup$ You are asking for $E5XY\mid Z)-E(X\mid Z)E(Y\mid Z)$ where $(X,Y,Z)$ is centered normal with some covariance matrix $C$. Surely you know how to compute at least some of these terms? $\endgroup$ – Did Jun 20 '17 at 13:00
  • $\begingroup$ Hello, Did. I have managed to compute $E[X|Z]$ and $E[Y|Z]$ under an assumption that $X, Z$ and $Y, Z$ are pairwise bivariate normal. But I have troubles to compute $E[XY|Z]$, since i am not sure, whether $XY$ and $Z$ are also bivariate normal and whether the $Cov(XY,Z) = 0$. I would very appreciate, if you cound provide any suggestion about it, $\endgroup$ – Tata Jun 20 '17 at 16:39
  • $\begingroup$ Well, in full generality, $X=aZ+bU$ and $Y=cZ+dU+eV$ for $(U,V)$ standard normal independent of $Z$ and some well chosen $(a,b,c,d,e)$, hence $E(XY\mid Z)=acZ^2+bd$. (No, $XY$ is not normal indeed...) $\endgroup$ – Did Jun 20 '17 at 17:03
  • $\begingroup$ Dear Did, thank you for your reply. However, I didn't get your idea completely. Do you mean, I should represent $X$ and $Y$ like you suggest and in this way obtain a result? If so, could you please suggest, what could the $U$ and $V$ look like? Also I updated my post with computation steps which I did. Would you be so kind to take a look at it and give me some comments? Thank you. $\endgroup$ – Tata Jun 20 '17 at 19:12
  • $\begingroup$ "Do you mean..." Indeed this is what I wrote (not "suggested"). "If so, could you please suggest, what could the U and V look like?" Already done, see previous comment. $\endgroup$ – Did Jun 20 '17 at 20:02

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