Solving a Multiplicative Stochastic Differential Equation with Itô's Lemma

Summary

The document solves a stochastic differential equation in which the level of a process is multiplied by a constant coefficient and a Brownian increment. It identifies the equation as a geometric Brownian motion with zero drift and a negative diffusion coefficient, then applies Itô's lemma to the logarithm of the process. The second derivative of the logarithm contributes a positive deterministic drift term, while the Brownian term retains the negative coefficient.

Integrating the resulting equation for the log process and exponentiating gives the stated solution in terms of the initial value, elapsed time, and Brownian motion. This example illustrates why directly integrating the original noisy differential misses the correction that appears under a nonlinear transformation. The derivation assumes a positive initial value so that the logarithm is defined, and it concerns a constant coefficient model; it does not discuss estimation or financial interpretation of the parameters.

Key ideas

• Applying Itô's lemma to the logarithm converts a multiplicative SDE into an additive one.
• The curvature term from the logarithm creates a deterministic drift in the log process.
• A negative diffusion coefficient changes the sign of the Brownian term without changing its variance contribution.
• The stated solution follows by integrating the log process and exponentiating.

Tags

# What is the SDE of this equation?


# What is the SDE of this equation?












I am new and struggling to understand how to solve this using Ito lemma.

Can someone please explain it to me:

$$dS_t=-\frac{1}{2}\sigma^2 S_t dW_t$$

what is the solution with explanation please

## Answer by Leguan3000 (score 3)

https://quant.stackexchange.com/a/50776

Actually this is just the Black-Scholes SDE with zero drift and $-\frac{1}{2}\sigma^2$ volatility. If you plug that into the well known solution, you get $S_t=S_0e^{\frac{1}{8}\sigma^4t-\frac{1}{2}\sigma^2 W_t}$ but let's calculate it with Ito's formula.

Choose $f(x)=\log(x)$, then we have $f'(x)=\frac{1}{x}$ and $f''(x)=-\frac{1}{x^2}$. Inserting in Ito's formula yields $$ d\log(S_t)=\frac{1}{S_t}dS_t+\frac{1}{2}\left(-\frac{1}{S_t^2}\right)d\langle S\rangle_t \\ =-\frac{1}{S_t}\frac{1}{2}\sigma^2S_tdW_t+\frac{1}{2}\frac{1}{S_t^2}\frac{1}{4}\sigma^4S_t^2dt \\ = \frac{1}{8}\sigma^4dt-\frac{1}{2}\sigma^2dW_t $$ or equivalently $$ \log(S_t)=\log(S_0)+\frac{1}{8}\sigma^4 t-\frac{1}{2}\sigma^2W_t \\ \Leftrightarrow S_t=S_0\exp\left(\frac{1}{8}\sigma^4t-\frac{1}{2}\sigma^2W_t\right) $$

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