Computing expectation of Poisson process which involves covariance

Question

I am looking at a past test with solutions, for question 8a, one of the step doesn't make sense to me. $$ \begin {aligned} E[(2 + B_2 + B_3)^2] &= E[4 + B_2^2 + B_3^2 + 4B_2 + 4B_3 + 2B_2B_3] \\ &= 4 + Var(B_2) +Var(B_3) + 4E(B_2) + 4E(B_3) + 2Cov(B_2, B_3) \end {aligned} $$ for me, I understand that $E(B_2^2) = Var(B_2)+E^2(B_2)$ and $E(B_3^2) = Var(B_3)+E^2(B_3)$, but I don't see how $E^2(B_2)+E^2(B_3)+2E[B_2B_3] = 2Cov(B_2,B_3)$, I tried looking up the definition of covariance but nothing's clicking, can anyone direct me towards how I should understand this step? thank you!

Answer 1

It's not. The unwritten property that is being used here is that $\operatorname{E}[B_i] = 0$, thus $$\operatorname{E}[B_i^2] = \operatorname{Var}[B_i] + \operatorname{E}[B_i]^2 = \operatorname{Var}[B_i]$$ and $$\operatorname{Cov}[B_i, B_j] = \operatorname{E}[B_i B_j] - \operatorname{E}[B_i]\operatorname{E}[B_j] = \operatorname{E}[B_i B_j].$$ You can see how in the next step, the solution evaluates $\operatorname{E}[B_2] = 0$ and $\operatorname{E}[B_3] = 0$.

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For the sake of completeness, I will quote the question below:

Let $\{B_t\}_{t \ge 0}$ be standard Brownian motion, and let $\tau = \inf \{t > 0 : B_t = -2 \text{ or } 3\}$. Compute $\operatorname{E}[(2+B_2+B_3)^2]$.

The fact that $\{B_t\}$ is standard Brownian motion is what justifies the claim $\operatorname{E}[B_t] = 0$. The stopping time $\tau$ is not used in this part of the question but in a later part which I have not quoted here because it is not part of the OP's question.