# Variance of combination of Brownian Motions

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.

$$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}$$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.
• 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 $T-t$? – rs4rs35 Mar 17 at 11:00