Product of correlated brownian motions

Question

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 $dB_xdB_y$ = $\rho dt$

Answer 1

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