stochastic calculus - How to express the volatility of two correlated Ito processes $Wt_1, Wt_2$ expressed in terms of $W_t$?

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 ?

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

Answer

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.