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Having two correlated Ito processes ($W_t^1$ and $W_t^2$ are correlated Brownian motions with correlation $\rho$)

$dX_{t} =\mu_{1} dt + \sigma_1 dWt_1 $

$dY_{t} = \mu_{2} dt + \sigma_2 dWt_2 $

How can the below be proven algebraically ?

$\sqrt{\sigma_1^2 + \sigma_2^2 +2 \sigma_1 \sigma_2 \rho} \ \ dW_t = \sigma_1 dW_t^1 + \sigma_2 dW_t^2$

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  • $\begingroup$ Are you trying to express it in terms of a third Brownian motion or in term of either one of $W^1$ or $W^2$? $\endgroup$
    – SRKX
    Commented Apr 26, 2016 at 4:25

1 Answer 1

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What can be shown is that the above expressions are equal in probability. First check the distribution. As any linear combination of a Gaussian is Gaussian the right hand side is Gaussian - the left hand side too. Then we need the 2 moments:

The expected values - it is zero ... easy to see.

Next what you did not specify is that the correlation between $dW_t^1$ and $dW_t^2$ is $\rho$ then the variance can be calculated by $$ VAR[\sigma_1dW_t^1+\sigma_2dW_t^2] = \sigma_1^2 VAR[dW_t^1] + 2 \sigma_1 \sigma_2 Covar[dW_t^1,dW_t^2] + \sigma_2^2 VAR[dW_t^2] $$ which equals $$ \sigma_1^2 dt + 2 \sigma_1 \sigma_2 \rho dt + \sigma_2^2 dt. $$

On the other hand the variance of the lhs: $$ VAR[\sqrt{\sigma_1^2 + 2 \sigma_1 \sigma_2 \rho+ \sigma_2^2} dW_t] = (\sigma_1^2 + 2 \sigma_1 \sigma_2 \rho+ \sigma_2^2) VAR[dW_t] $$ and this is $$ (\sigma_1^2 + 2 \sigma_1 \sigma_2 \rho + \sigma_2^2) dt, $$ exactly what we needed.

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    $\begingroup$ Nice answer but just a pedantic remark: $\text{VAR}[dW_t^i] = dt$ not $t$, same point for covariance. $\endgroup$
    – Quantuple
    Commented Apr 26, 2016 at 7:42
  • $\begingroup$ I have changed this just some seconds ago ;) $\endgroup$
    – Richi Wa
    Commented Apr 26, 2016 at 7:43
  • $\begingroup$ Wow that was fast ;) $\endgroup$
    – Quantuple
    Commented Apr 26, 2016 at 7:43
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    $\begingroup$ @Quantuple Also, pedantically : $\textrm{VAR}[dW_t^i]$ has no mathematical meaning. ;-) $\endgroup$
    – Olórin
    Commented Apr 14, 2019 at 18:22
  • $\begingroup$ the dXYZ notion is not perfect but you get used to it, right? ;) $\endgroup$
    – Richi Wa
    Commented Apr 15, 2019 at 18:05

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