Applying Itô’s Product Rule to Correlated Geometric Brownian Motions

Summary

The document asks how to derive the stochastic differential equation for a portfolio quantity defined as twice the product of two correlated stock prices. Each stock follows geometric Brownian motion with its own drift and volatility, and the question focuses on the cross variation term that appears when applying Itô’s product rule.

For the product of the stocks, the quadratic covariation is driven by their diffusion terms: it is proportional to the product of their volatilities, their instantaneous correlation, and the product of their current prices over an infinitesimal time interval. Dividing the resulting differential by the portfolio value gives the relative SDE; multiplying the product by two does not change its relative dynamics. The supplied document contains only the question and no answer, so it does not show a full derivation or resolve its notation issues.

Key ideas

• Itô’s product rule adds a quadratic covariation term when multiplying two stochastic processes.
• For correlated geometric Brownian motions, the cross term depends on both volatilities and their correlation.
• The factor of two in the defined portfolio scales its level but cancels from its relative dynamics.
• The document raises the derivation question but provides no worked solution.

Tags

# SDE for a portfolio of two correlated assets $ Y_{t} = 2 S^{1}_{t} S^{2}_{t}$


# SDE for a portfolio of two correlated assets $ Y_{t} = 2 S^{1}_{t} S^{2}_{t}$












I am analysing a problem where I have two correlated stocks described by Brownian motions $$ \frac{dS^{1}_{t}}{S^{1}_{t}}=\mu_{1} dt + \sigma_{1} dW^{1}_{t} \quad \quad (1)$$ $$ \frac{dS^{2}_{t}}{S^{2}_{t}}=\mu_{2} dt + \sigma_{2} dW^{2}_{t} \quad \quad (2)$$

where the $$ Y_{t} = 2 S^{1}_{t} S^{2}_{t} \quad \quad (3)$$

I am looking for SDE to this equation. I know that given the below two geometric Brownian motions $$ dX^{1}_{t} =z_{1} dt + Y_{1} dBt \quad \quad (4)$$ $$ dX^{2}_{t} =z_{2} dt + Y_{2} dBt\quad \quad (5)$$

the below equality can be derived

$$d(x^{1}_{t} x^{2}_{t})=x^{1}_{t} dx^{2}_{t} + x^{2}_{t} dx^{1}_{t} + Y_1 Y_2 dt \quad \quad (6)$$

I know as well that the two Brownian motions can be presented as $$ W_t = \rho W^{1}_{t} + \sqrt{1-\rho^2} W^{2}_{t} \quad \quad (7)$$

However when I try to apply the properties to the above problem I am getting confused.

The manual says that $$\frac{dY_t}{Y_t} = 2 \big{(} \frac{dS^{1}_{t}}{S^{1}_{t}} + \frac{dS^{2}_{t}}{S^{2}_{t}} + \frac{ <S^{1} S^{2}>_t}{S^{1}_{t} S^{1}_{t}} \big{)} \quad \quad (8)$$

Can anybody clarify how the $<S^{1} S^{2}>_t$ is derived? how this relates to the $Y_1 Y_2 dt$ from the eq.(6) ? why sigmas were not used analogically ? and what it should be substituted for in the further calculation?

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