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Consider that the correlation between two standard brownian motions $dB_x$ and $dB_y$ be $\rho$. And we write $\mathtt{Cor} (dB_x,dB_y)$ = $\rho$. Show that $dB_xdB_y$ = $\rho dt$

Carlos Lobato
  • 139
  • Any chance you might add something personal to this question, like, dunno, what you tried to solve it? – Did May 14 '13 at 17:03
  • I started by the definition $Cor(dBx,dBy) = Cov(dBx,dBy)/dt$, because $Var(dB_x)=dt$ and $E(dB_x=0)$ so I get $E(dBxdBy) = \rho dt$, but I don't know hot to go on – Carlos Lobato May 15 '13 at 15:55

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I solved and one should use these rules:

.$\left(dt\right)^{n}\in$o$\left(dt\right)$

.$\mathbb{E}[f(B_t)]=f(B_t)$ if $Var[f(B_t)]=0$

.$dB_t^2=dt$

From the result that I already have i.e.

$\mathbb{E}(dBxdBy)=\rho dt$

We just have to show that $Var(dBxdBy)=0$, and apply the second rule, but

$Var(dBxdBy)=\mathbb{E}(dBxdBy)^2-\mathbb{E}^2(dBxdBy)=\mathbb{E}(dtdt)-\rho2dt^2=dt^2-\rho^2dt^2=0$

QED

Carlos Lobato
  • 139