Show that for $B = (B_t)$ Brownian motion, its covariance is $cov(B_s, B_t) = min(s, t)$.

The solution I was given was:

For $s ≤ t$, $B_t = B_s + (B_t − B_s)$,

$B_sB_t = B_s^2 + Bs(Bt − Bs)$

$cov(B_s,B_t)=E[B_sB_t]$(as $E(B_i)=0)$

so $cov(B_s,B_t)=E[B_s^2]+E[B_s(B_t-B_s)]$,

as all increments of Brown motion are independent the second term in the RHS

$=E[B_s]E[B_t-B_s]=0*0=0$. Now $cov(B_s,B_t)=E(B_s^2)=Var(B_s)=s$.

Similarly if $t\leq s$ we get $=t$.

My question is, could I not have done this arguement the same way without assuming that $s\leq t$ in the first place? I mean, if $s\leq t$ why cant I say $B_s = B_t + (B_s − B_t)$ and conitnue like this to get the answer with a max instead of a min?

Is it because $B_t-B_s$ is only an increment if $t\geq s$? I mean otherwise we dont have independency or something?

  • 3
    $\begingroup$ Why don't you first assume $s \leq t$ without loss of generality? $\endgroup$ – Calculon May 8 '15 at 21:58

In your proof you use the fact that $B_s$ is independent of $B_t-B_s$ (indepedent increments) but this is only true if $s\leq t$.

  • $\begingroup$ thank you thats what I thought too $\endgroup$ – vounoo May 8 '15 at 21:56

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