Deriving the Generator PDE for a Geometric Brownian Motion Expectation

Summary

The document asks which partial differential equation is satisfied by an expectation of a function evaluated at a Wiener-driven, exponentially transformed variable. The proposed derivation differentiates the expectation and applies Itô’s rule to the exponential process, observing that its stochastic differential has a Brownian increment after the drift terms cancel. The question then asks how this leads to a PDE.

The answer identifies the infinitesimal generator as one half of the state squared times the second spatial derivative, corresponding to a diffusion with no drift. This generator provides the spatial operator for the associated backward equation, though the response itself is very brief and does not work through the full PDE or clarify the time variable and terminal condition. The document is a compact stochastic-calculus prompt rather than a complete derivation, so readers should verify the precise sign and time convention for the PDE they need.

Key ideas

• Applying Itô’s rule to the exponential Wiener process cancels its drift terms in the stated setup.
• The response identifies the diffusion generator as one half of the squared state multiplied by the second spatial derivative.
• A generator supplies the spatial operator for a corresponding backward equation.
• The brief answer does not establish the full PDE, time convention, or boundary conditions.

Tags

# Which PDE is satisfied by the function of Wiener process $u(t,x)$?


# Which PDE is satisfied by the function of Wiener process $u(t,x)$?












Suppose you have the following function:

$u(t,x)=\mathbb{E}[f(xe^{W_t+\frac{1}{2}t})]$, where $W_t$ is a Wiener process.

Let us first differentiate:

$du=\mathbb{E}[f'(xe^{W_t+\frac{1}{2}t})(e^{W_t-\frac{1}{2}t}dx+xd(e^{W_t-\frac{1}{2}t}))]$ and using some properties of the Brownian motion

$d(e^{W_t-\frac{1}{2}t})=e^{W_t-\frac{1}{2}t}(dW_t-\frac{1}{2}dt+\frac{1}{2}(dW_t)^2)=e^{W_t-\frac{1}{2}t}dW_t$

Clearly, $\mathbb{E}[f'(xe^{W_t+\frac{1}{2}t})xd(e^{W_t-\frac{1}{2}t})dW_t]=0$

and, therefore, $du=\mathbb{E}[e^{W_t-\frac{1}{2}t}f'(xe^{W_t+\frac{1}{2}t})]dx+0 dt$

Could someone tell me from this how to find the PDE that the function u satisfies?

## Answer by FunnyBuzer (score 2, accepted)

https://quant.stackexchange.com/a/39969

Since we have $\frac{\partial\mathbb{E}[u(t,x)]}{\partial t}=0$, the infinitesimal generator is:

$Af(x)=\frac{1}{2}x^2\frac{\partial}{\partial x^2}$

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