Autocovariance of Itō Process Increments at Different Times

Summary

The document asks whether increments of an Itō process are uncorrelated or independent at distinct times, given an adapted drift and volatility. The response attempts to expand the product of increments into drift and Brownian-motion terms, then concludes that the expectation vanishes by treating Brownian increments at different times as effectively the same increment.

That reasoning is not reliable: increments at distinct times are not interchangeable, and zero covariance would not by itself establish independence. The answer does not correctly resolve the role of adapted drift and volatility, so it should not be used as a derivation. Its value is in surfacing a useful distinction for stochastic modeling: one must specify the process and the time intervals, and separately assess covariance and independence rather than infer one from the other.

Key ideas

• The question distinguishes zero autocovariance from independence of increments.
• Brownian increments over distinct nonoverlapping intervals are not the same random variable.
• A zero covariance result alone does not imply that increments are independent.
• The response’s conclusion relies on an invalid interchange of Brownian increments and does not resolve the general Itō-process case.

Tags

# Autocovariance of increments of a semimartingale


# Autocovariance of increments of a semimartingale












Say that $X_t$ is an Itō process with \begin{equation} dX_t = \mu_t dt + \sigma_t dW_t \end{equation} where $\mu_t$ and $\sigma_t$ are adapted processes.

Is it always true that \begin{equation} E[dX_t dX_s] = 0, \quad t\neq s \end{equation} i.e. the increments of $X_t$ are always independent? At first glance I would say yes since $dX_t dX_s \propto dtds$ which will integrate to zero, but is there something more that can be said about $dX_t dX_s$?

## Answer by numerairX (score 1)

https://quant.stackexchange.com/a/42927

**please correct me if the math is wrong!!

I think upon breaking down the products $E(dX_tdX_s)$, we have the $dtds$, $dtdW_s$ terms which all turns out to be 0. It leaves $E(dW_tdW_s)$ which comes down to sharing the same wiener process $dW = \sqrt{dt}Z$, where Z follows N(0,1).

However note that since we're calculating expecation, $dW_t$ and $dW_s$ is basically the same thing under integration, because they are both denoting a change in dt. so $E(dW_tdW_s)$ is the same as $E(dW_tdW_t)$, which equals 0.

Hence it seems like the equation always equals to 0.

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