Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Do you really want to ask for the covariance of $W(t)$ and $0$? – Douglas Zare Apr 6 '11 at 6:33
Again? – Did Apr 6 '11 at 13:36
up vote 0 down vote accepted

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.

share|cite|improve this answer
So is what I did incorrect then? – icobes Apr 6 '11 at 18:34
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.