When Expectation Commutes with Differentiation in an SDE

Summary

The document asks when the expected value of an infinitesimal change in a stochastic process can be written as the differential of its expectation. The answer rewrites the stochastic differential equation in integral form: the process equals its initial value plus an accumulated drift integral and a stochastic integral. Under suitable conditions, the stochastic integral is a martingale with zero expected increment.

The remaining expectation is moved inside the drift integral using Fubini's theorem, yielding the expected drift as the time derivative of the process expectation. This justifies the shorthand equality between the expected differential and the differential of the expectation in this setting. The conclusion depends on regularity and integrability assumptions, including conditions that make the stochastic integral a martingale and permit exchanging expectation with integration. It is not an unconditional rule for arbitrary stochastic processes or coefficients.

Key ideas

• Write the stochastic differential equation as an integral equation before taking expectations.
• Under suitable conditions, the stochastic integral has zero expectation because it is a martingale.
• Fubini's theorem permits moving expectation inside the drift integral when its conditions hold.
• The resulting identity relies on assumptions about the process and integrability.

Tags

# Swapping expectation operator with differential operator


# Swapping expectation operator with differential operator












Suppose I have a general SDE

$dx_{t} = \mu dt + \sigma dz_{t}$

Then I can put $E[]$ on both sides to get

$E[dx_{t}] = E[\mu dt] + E[\sigma dz_{t}]$

Now comes the question: I've seen some formulas where

$E[dx_{t}]$ becomes $dE[x_{t}]$

Is it ok to swap $E[.]$ with $d[.]$?

Thank you.

## Answer by AFK (score 1)

https://quant.stackexchange.com/a/16293

Remember that $dx_t = \mu_t dt + \sigma_t dz_t$ is just a shorter notation for $$ x_t = x_0 + \int_0^t \mu_s ds + \int_0^t \sigma_s dz_s $$

Now, under mild hyopthesis on $\sigma$ the stochastic integral is a martingale so $E[\int_0^t \sigma_s dz_s] = E[\int_0^0 \sigma_s dz_s] = 0$. We are left with $$ E[x_t] = x_0 + E[\int_0^t \mu_s ds] = x_0 + \int_0^t E[\mu_s] ds $$ by Fubini's theorem.

So $dE[x_t] = E[\mu_t] dt$. This justifies writing $dE[x_t] = E[dx_t]$.

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