Why Ito Process Moments Require Integrating the Drift and Diffusion

Summary

The document corrects an attempt to infer the mean and variance of an Ito process by multiplying its instantaneous drift and variance rates by elapsed time. An Ito differential is shorthand for an integral over the path: the state at time t equals its initial value plus accumulated drift and a stochastic integral. Taking expectations removes the stochastic integral under the stated assumptions, leaving the initial value plus the time integral of the drift along the process.

The proposed mean formula follows only when the initial state is zero and the drift is constant. The response asks the reader to examine the second moment to determine the variance, but does not complete that derivation. In general, the drift and diffusion coefficients may depend on time and the evolving state, so their effects cannot simply be evaluated once and multiplied by time. The note therefore establishes the integral reasoning for the mean while leaving the variance calculation incomplete.

Key ideas

• An Ito differential represents an integrated process over time, not a finite change with fixed coefficients.
• The expected state is the initial value plus the time integral of the drift along the process.
• Multiplying drift by elapsed time gives the mean only under restrictive initial-value and constant-drift assumptions.
• The document does not finish the variance derivation, which requires analyzing the second moment of the integrated process.

Tags

# Can the differential operator be removed to get the mean/variance of an Ito process?


# Can the differential operator be removed to get the mean/variance of an Ito process?












If $X_t$ is an Ito process, such that:

$dX_t = \mu(t, X_t)dt + \sigma(t, Xt)dW_t$ where $W_t$ is a standard brownian motion.

Then we can say that:

$E(dX_t) = \mu(t, X_t)dt$ and that $Var(dX_t) = \sigma^2(t, Xt)dt$

Is this equivalent to saying that (I removed the differential operator):

$E(X_t) = \mu(t, X_t)\times t$ and that $Var(X_t) = \sigma^2(t, Xt)\times t$

## Answer by fni (score 4, accepted)

https://quant.stackexchange.com/a/22149

This is wrong! Notice that $dX_t=\mu(t,X_t)dt + \sigma(t,X_t)dW$ is a shorthand for $$\int_0^tdX_s = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s$$ Integrating: $$X_t-X_0 = \int_0^t \mu(s,X_s)ds + \int_0^t\sigma(s,X_s)dW_s \text{ (eq.1)} $$ If we take expectations, remembering that $\mathbb{E}[\int_0^t\sigma(s,X_s)dW_s]=0$, we have $$\mathbb{E}[X_t]=X_0 + \int_0^t \mu(s,X_s)ds$$ To have your result, we need $X_0=0$ and constant drift $\mu(t,X_t)=\mu$. You compute the second moment of (eq.1) to check what happens to the variance

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