# covariance function for Brownian motion

What is Cov[W(t),W(0)] when W(t) is t*B(1/t) and W(0) = 0 where B(t) is standard Brownian motion. The answer is min {s,t}. I am unsure how they get that because I get min{0,1}. Here is what I did:

Cov[W(t),W(0)] = E[W(t)W(0)] = E[t*B(1/t)*B(0)] = tE[B(1/t)*B(0)]

If t <= s, then the expectation is t*(1/t) = 1 and is s>= t then the expectation is t*0 = 0.

Thanks for the help!

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Do you really want to ask for the covariance of $W(t)$ and $0$? –  Douglas Zare Apr 6 '11 at 6:33
–  Did Apr 6 '11 at 13:36

That answer, obviously, corresponds to a different exercise, which can be formulated as follows: What is ${\rm Cov}(W(s),W(t))$, $0 \leq s \leq t$, when $W(t)=t B(1/t)$ and $W(0)=0$, where $B$ is a standard BM. [The point in defining $W(0)=0$ is that $1/0$ is not defined; further, note that this definition agrees with the variance of $W(t)$ as $t \downarrow 0$.]
General remark: by symmetry, it suffices to calculate covariance functions for $s \leq t$.
EDIT: The purpose of this exercise is to show that by a simple transformation, namely $W(t)=t B(1/t)$, one BM can be transformed into another. Note that the processes $W:=\{W(t):t \geq 0\}$ and $B:=\{B(t):t \geq 0\}$ are equal only at two points, $t=0$ and $t=1$, yet they have exactly the same law (both are standard BM). [Note that the law of a mean zero Gaussian process is completely determined by its covariance function; the fact that $W$ is a mean zero Gaussian process follows straightforwardly from the fact that $B$ is.]
So, you want to show that $W$ has the same covariance function as $B$. Recall that, by symmetry, it suffices to calculate the covariance function for $0 \leq s \leq t$. In particular, letting $s=0$, you want to show that ${\rm Cov}(W(0),W(t)) = \min\{0,t\}$, for any $t \geq 0$. This is obvious.
@icobes: Indeed. Note that since $W(0)$ is constant, $W(0)$ and $W(t)$ are independent. So, ${\rm Cov}(W(0),W(t)) =$? –  Shai Covo Apr 6 '11 at 19:26