1
$\begingroup$

Let $Z(t)=W(t)-\frac{t}{T}W(T-t)$ for any $0\leq t\leq T$ with $W(t)$ a Brownian motion, find the variance of $Z(t)$.

My attempt:

$Var(Z(t))=\mathbb{E}(Z(t)^{2})-\mathbb{E}(Z(t))^{2}$

$Z(t)=W(t)-\frac{t}{T}W(T-t)=W(t)-\frac{t}{T}(W(T-t)-W(T))+\frac{t}{T}W(T)$

Here $W(T-t)-W(T)$ is a reflected Brownian motion going from $T$ to $0$.

Also, $\frac{t}{T}W(T)=\frac{T}{t}\frac{t}{T}W(T\frac{t}{T})=W(t)$ by the scaling property of Brownian motion.

Now, according to my calculations $\mathbb{E}(Z(t)^{2})-\mathbb{E}(Z(t))^{2}=8t+\frac{t^{3}}{T^{2}}-4\mathbb{E}(\frac{t}{T}W(t)(W(T-t)-W(T)))$.

Here I get stuck, since I cannot conclude that the Brownian motions are independent, since for $t=\frac{T}{2}$ they are equal.

$\endgroup$
1
$\begingroup$

$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$Hint: You can find $\Var(Z(t))$ by noting that in general, we have $$\Var(aW(t) + bW(s)) = a^2 \Var(W(t)) + 2ab\Cov(W(t), W(s)) + b^2 \Var(W(s)).$$ I'm assuming you know what $\Cov(W(t), W(s))$ equals. If not, you could see here: Find the covariance of a brownian motion.

$\endgroup$
  • $\begingroup$ Thank you, but that means the expression (for clarification) must be split up for $t>T-t$ and $t<T-t$, since $t$ is not always greater or smaller than $T-t$? $\endgroup$ – rs4rs35 Mar 17 at 11:00
  • $\begingroup$ Pretty much, yes. $\endgroup$ – Minus One-Twelfth Mar 17 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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