Deriving Uncorrelated Combinations of Correlated Geometric Brownian Motions
Summary
The document shows how to establish that two weighted combinations of log prices from correlated stochastic processes are uncorrelated. It starts with correlated geometric Brownian motions, takes logarithms, and forms a sum and difference using the assets’ volatility parameters. The derivation computes each component’s variance and their covariance over time.
The key result follows because the weighted log-price terms have equal variances, while their covariance is proportional to the processes’ correlation. In the sum and difference, the cross-covariance terms cancel, yielding zero covariance; the variances scale with one plus or minus the correlation. The derivation assumes the stated diffusion model and constant parameters. It also notes that zero correlation implies independence when the variables are jointly normally distributed, a qualification that matters when applying the result beyond this setting.
Key ideas
- Taking logarithms turns the diffusion’s random component into a scaled Brownian motion.
- The weighted log-price terms have equal variances under the model assumptions.
- The sum and difference have variances proportional to one plus and one minus the correlation, respectively.
- Their covariance is zero because the cross terms cancel.
- For jointly normal variables, zero correlation also implies independence.
Tags
Full text
# Correlated Stochastic Processes
# Correlated Stochastic Processes
Let say, I have 2 stochastic processes: $$\begin{align} dS_1 &= \left( r - q_1 \right)S_1 dt + \sigma_1 S_1 dW_1 \\ dS_2 &= \left( r - q_2 \right)S_2 dt + \sigma_2 S_2 dW_2 \end{align}$$ The correlation between these 2 processes is $\rho$. Now I define 2 new processes as: $$\begin{align} x_1 = \sigma_1 \log S_2 + \sigma_2 \log S_1 \\ x_2 = \sigma_1 \log S_2 - \sigma_2 \log S_1 \end{align}$$
As per Hull's book, these processes $x_1, x_2$ are uncorrelated with standard deviation $\sigma_1 \sigma_2 \sqrt{2 \left( 1+\rho \right)}$ and $\sigma_1 \sigma_2 \sqrt{2 \left( 1-\rho \right)}$ respectively.
How can I show this result?
## Answer by Daneel Olivaw (score 3)
https://quant.stackexchange.com/a/59113
I have the impression the expressions in your question miss the time term $t$, though this does not change much. Define for $i,j\in\{1,2\}$: $$\begin{align} s_i(t)&=\log S_i(t) \\ y_{i,j}(t)&=\sigma_is_j(t) \end{align}$$ Then: $$\begin{align} V(y_{i,j}(t))&=\sigma_i^2V(s_j(t)) \\ &=\sigma_i^2V(\sigma_jW_j(t)) \\ &=\sigma_i^2\sigma_j^2t \\ &=V(y_{j,i}(t)) \end{align}$$ Moreover: $$\begin{align} C(y_{1,2}(t),y_{2,1}(t))&=\sigma_1\sigma_2C(s_2(t),s_1(t)) \\ &=\sigma_1^2\sigma_2^2C(W_2(t),W_1(t)) \\ &=\sigma_1^2\sigma_2^2\rho t \end{align}$$ Hence: $$\begin{align} &V(x_1(t))=\sigma_1^2\sigma^2_2t+\sigma_2^2\sigma_1^2t+2\sigma^2_1\sigma^2_2\rho t=\sigma_1^2\sigma_2^22(1+\rho)t \\ &V(x_2(t))=\sigma_2^2\sigma^2_1t+\sigma_1^2\sigma_2^2t-2\sigma^2_1\sigma^2_2\rho t=\sigma_1^2\sigma_2^22(1-\rho)t \end{align}$$ Finally, by bi-linearity and symmetry of covariance: $$\begin{align} C(x_1(t),x_2(t))&=C(y_{1,2}(t)+y_{2,1}(t),y_{1,2}(t)-y_{2,1}(t)) \\ &=V(y_{1,2}(t))-C(y_{1,2}(t),y_{2,1}(t))+C(y_{2,1}(t),y_{1,2}(t))-V(y_{2,1}(t)) \\ &=V(y_{1,2}(t))-V(y_{2,1}(t)) \\ &=0 \end{align}$$ $x_1$ and $x_2$ are uncorrelated $-$ note that for normally-distributed random variables, null correlation also implies independence.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.