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:



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.


$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$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.

  • $\begingroup$ 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

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