Product Rule for Multiplying Stochastic Exponentials

Summary

The document shows how to establish a product identity for stochastic exponentials of continuous semimartingales. It starts with processes whose stochastic exponentials solve differential equations of the form dZ = Z dX. Applying the semimartingale product rule to the two solutions introduces a quadratic covariation term. The resulting differential equation has the same form as the stochastic exponential driven by the sum of the original processes and their covariation.

By identifying the product as the solution to that equation, the answer obtains the identity that the product of the two stochastic exponentials equals the stochastic exponential of the combined process. The argument relies on the continuous-semimartingale setting and the stated initial values. It is a concise proof sketch, rather than a treatment of discontinuous processes, existence conditions, or applications in financial modeling.

Key ideas

• Each stochastic exponential is defined as the solution of a stochastic differential equation driven by its underlying process.
• The semimartingale product rule adds a quadratic covariation term when multiplying the solutions.
• The resulting differential equation matches the stochastic exponential driven by the sum plus covariation.
• The proof is stated for continuous semimartingales with zero initial values.

Tags

# Properties of Stochastic Exponential


# Properties of Stochastic Exponential












Let $\{X_t\}_{t \ge 0},\{Y_t\}_{t \ge 0}$ be a continuous semi-martingale with $X_0 = Y_0 = 0$, let ${\cal E}(X)$ to be the unique solution of: $dZ_t = Z_t dX_t$ with $Z_0=1$.

We can show that ${\cal E}(X)_t = exp(X_t - \frac{1}{2}[X]_t)$, but how to show that ${\cal E}(X){\cal E}(Y) = {\cal E}(X+Y+[X,Y])$ where $[X，Y]$ denotes the quadratic covariation between $X_t$ and $Y_t$.

## Answer by Calculon (score 1)

https://quant.stackexchange.com/a/39122

Let $dV_t = V_tdY_t$. We will consider $d(VZ)_t$. \begin{align} d(VZ)_t &= V_tdZ_t + Z_tdV_t + d[V,Z]_t \\ &= V_tZ_tdX_t + Z_tV_tdY_t + Z_tV_td[X,Y]_t \\ & = V_tZ_td(X + Y + [X,Y])_t \end{align} I used the product rule above and also the fact that stochastic integral is linear in the integrator.

We have $V_tZ_t = \mathcal{E}(X)\mathcal{E}(Y)$. On the other hand, $V_tZ_t = \mathcal{E}(X+Y+[X,Y])$ by the SDE above.

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