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We have Brownian motion $X_t=at+oB_t$, where $B_t$ represents a standard Brownian motion. To compute expectation, I do:

$E[X_t]=E[at+oB_t]=at+oE[B_t]$

Since a standard Brownian motion is normally distributed with mean 0 and variance 1, $E[X_t]=at$

For variance,

$Var[X_t]=E[X_t]^2-(E[X_t])^2=(at)^2-E[at+oB_t]^2=(at)^2-(a^2t^2+E[2atoB_t]+E[oB_t]^2=(at)^2-o^2E[B_t]^2$.

Since $E[B_t]=0$, then $Var[X_t]=(at)^2-(a^2t^2)=0$

Which doesn't make sense. Where did I go wrong?

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    $\begingroup$ $Var[X_t] = E[X_t^2] - (E[X_t])^2$, my friend. $\endgroup$ – Jane Maths Feb 3 '17 at 15:10
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Firstly, as Jane said, $\mathsf {Var}(Z)=\mathsf E(Z^2)-\mathsf E(Z)^2$ and the very point is that, except for degenerate random variables, $\mathsf E(Z^2)> \mathsf E(Z)^2$ .

Secondly, you may be better off using the same tactic, since you know $\mathsf {Var}(B_t)=1$ because $B_t\sim\mathcal{N}(0,1^2)$

So   $\mathsf {Var}(X_t)~=~\mathsf {Var}(at+oB_t)~=~ \ldots$

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