Combining Correlated Brownian Risk into One Volatility

Summary

The document derives the variance of a weighted sum of two correlated Brownian increments. For coefficients applied to each increment, the variance includes both individual variance terms and a covariance term determined by their correlation. Since each Brownian increment has variance equal to the time step and covariance equals correlation times that step, the combined variance is the time step multiplied by the sum of the squared coefficients and twice their correlation-weighted product.

A single Brownian increment scaled by the square root of that coefficient expression has matching mean and variance. The answer argues that the two quantities have the same Gaussian distribution, so the equality is distributional rather than an algebraic identity for particular sample paths. This gives an effective volatility for aggregating correlated sources of risk. The explanation assumes constant coefficients, Gaussian Brownian increments, and a valid correlation; it does not address time-varying dependence or more than two processes.

Key ideas

• The variance of a weighted sum of Brownian increments includes their covariance as well as each individual variance.
• Correlation determines the cross term in the effective variance of the combined process.
• A single scaled Brownian increment can match the weighted sum’s Gaussian distribution through its variance.
• The equivalence is in distribution and does not claim pathwise equality for a chosen Brownian realization.

Tags

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


# 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 by Richi Wa (score 5, accepted)

https://quant.stackexchange.com/a/25642

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.