If $W_t$ is a Wiener process, what is $ \operatorname{Var}(W_t+W_s)$?

Question

Doing some questions from the book '150 Most Frequently Asked Questions on Quant Interviews', I try to solve the following: If $W_t$ is a Wiener process, what is $ \operatorname{Var}(W_t+W_s)$? It seemed pretty straightforward to me, and solved as follows, for $t$ and $s$ positive:

$\begin{align*} \operatorname{Var}(W_t+W_s) &= \operatorname{Var}(W_t)+ \operatorname{Var}(W_s)+2 \operatorname{Cov}(Wt,Ws)\\ &= E[W_t^2]-(E[W_t])^2+E[W_s^2]-(E[W_s])^2+2(E[W_tWs]-E[W_t]E[Ws])\\ &= E[W_t^2] + E[W_s^2] + 2E[W_tWs]\\ &= t+s+2\min(s,t) \end{align*}$

However, the solution in the book is: $ \operatorname{Var}(W_t+W_s) =\max(s,t)+3\min(s,t)$

Is my solution wrong? If yes, what part of it is wrong?

Answer 1

I was staring at this for a while before realizing your solutions are equivalent. For example, if $s < t$, then $$ \begin{aligned} t + s + 2 \min(s, t) &= \max(s, t) + \min(s, t) + 2 \min(s, t) \\ &= \max(s, t) + 3 \min(s, t). \end{aligned} $$

Answer 2

Your solution is correct - and so is theirs. This is because $$t+s+2\mathrm{min}(s,t) = \mathrm{max}(s,t) + 3\mathrm{min}(s,t).$$

What I'm wondering about is how they got to that form.

Answer 3

Nothing is wrong -- since $t+s = \max(s,t) + \min(s,t)$ your solution equals the book's.

It may simplify the calculation to assume WLOG that $t<s$ -- then $W_t+W_s=2W_t+(W_s-W_t)$, and $W_t$ and $W_s-W_t$ are independent.