I have a question about the variance of the following formula: $Var(W(t) - \frac{t}{T}W(T-t))$. Where $W(t)$ is a Brownian motion. I tried the following: $Var(W(t) - \frac{t}{T}W(T-t)) = E[W(t) - \frac{t}{T}W(T-t)]^2 - (E[W(t) - \frac{t}{T}W(T-t)])^2$. Now $E[W(t) - \frac{t}{T}W(T-t)] = E[W(t)] - \frac{t}{T}E[W(T-t) = 0 - \frac{t}{T}*T = t$, and:
$E[W(t) - \frac{t}{T}W(T-t)]^2 = E[W^2(t) - \frac{2t}{T}W(t)W(T-t) + \frac{t^2}{T^2}W^2(T-t)]$, this expectation consists of three terms:
$E[W^2(t)] = Var(W(t)) + E[W(t)]^2 = t + 0 = t$
$\frac{t^2}{T^2}E[W^2(T-t)] = \frac{t^2}{T^2}(Var(W^2(T-t)) + E[W(T-t)]^2) = \frac{t^2}{T^2}(t + T)$
$\frac{2t}{T}E[W(t)W(T-t)] = $ Here I'm getting stuck, I'm not sure if this is even the correct approach or that I should just completely change my way of calculating the variance.
Thanks in advance
Firstly, you have miscalculated the expectation. For any $s$, assuming that you consider a Brownian motion started from $0$, $\mathbb{E}[W(s)] = 0$. In particular, with $s = T-t$, we get that $$\mathbb{E}[W(T-t)] = 0$$ so that $$\mathbb{E}[W(t) - \frac{t}{T}W(T-t)] = 0 + \frac{t}{T} \cdot 0 = 0.$$
Now I move onto the variance. You have correctly found the $3$ expectations you'll need to compute for this and computed the first of those correctly. For the second, you should get that $$\frac{t^2}{T^2}E[W(T-t)^2] = \frac{t^2}{T^2} (T-t)$$ by using the fact that the covariance of $W$ is known to be given by $\mathbb{E}[W(t)W(s)] = t \wedge s$ with $t = s = T-t$.
Finally, for the third term again use the known of the covariance of Brownian motion to get that $$\frac{2t}{T}\mathbb{E}[W(t)W(T-t)] = \frac{2t}{T} (t \wedge (T-t))$$