Find the distribution of some random variable connected to Wiener Process. Please, check my solution.

Question

I need to find a distribution of $ 5W_1-W_3+W_7 $, where $W_t$ stands for Wiener Process $W_t\sim\mathcal{N}(0,t)$. Is this solution right? $E(5W_1-W_3+W_7)=5E(W_1)-E(W_3)+E(W_7)=0$ and since $W_7-W_3$ has the same distribution as $W_{7-3}=W_4 \sim \mathcal{N}(0,4)$

$D^2(5W_1-W_3+W_7)=D^2(5W_1+W_4)=5^2D^2(W_1)+D^2(W_4)+2\mathrm{cov}(5W_1,W_4)=25+4+2\cdot 5\mathrm{cov}(W_1,W_4)=29+10\mathrm{min}\{1,4\}=39$

Hence the final answer is $(5W_1-W_3+W_7) \sim\mathcal{N}(0,39)$

If something is wrong can you point it out for me, please? Thanks in advance!

Answer 1

Just write $W_t = \sum_{i=1}^t X_i$ where $X_i$'s are i.i.d. $\mathcal{N}(0, 1)$. You'll see that \begin{align} \text{Cov}(W_t, W_{t+\ell}) = \text{Var}(W_t) = t \end{align} for $\ell \ge 0$. From this, you can get \begin{align} \text{Var}(5W_1 - W_3 + W_7) & = \text{Var}(5W_1) + \text{Var}(W_3) + \text{Var}(W_7) + \\ & \phantom{{} = {}} - 2\text{Cov}(5W_1, W_3) + 2\text{Cov}(5W_1, W_7) - 2 \text{Cov}(W_3, W_7) \\ & = 25 + 3 + 7 - 10 + 10 - 6 = 29. \end{align}