Finding Expected Integrated Variance in a GARCH(1,1) Diffusion

Summary

The document asks how to compute the expected integrated variance over a one-unit interval when instantaneous variance follows a mean-reverting diffusion with GARCH(1,1)-style dynamics. It moves the expectation inside the time integral and then takes expectations of the stochastic differential equation. The stochastic integral has zero expectation under the usual integrability conditions, leaving an integral equation for expected variance that the author has not yet solved.

The material is useful as a setup for deriving the quantity, but it does not provide the requested closed-form expression or work through the final integration. In particular, it does not state initial conditions, parameter restrictions, or assumptions that ensure the stochastic integral is a martingale. A reader would need to solve the resulting first-order expectation equation and integrate its solution across the target interval. The displayed dynamics and time interval define the problem, but the document offers no numerical example, empirical evidence, or discussion of model limitations.

Key ideas

• Taking expectations removes the stochastic integral when the required integrability conditions hold.
• The expected variance satisfies a first-order integral equation under the stated mean-reverting diffusion.
• Expected integrated variance is obtained by integrating expected instantaneous variance over the interval.
• The document sets up the derivation but does not complete the solution.

Tags

# Expression for the expectation of Integrated variance in case of GARCH(1,1) process


# Expression for the expectation of Integrated variance in case of GARCH(1,1) process












I have the following SDE (GARCH(1,1)) for the instantaneous variance:

$$ d\sigma_t^2 = \kappa (\theta - \sigma_t^2) dt + \psi \sigma_t^2 dW_t $$

I would like to find an expression for $IV_t = E[\int_{t-1}^t \sigma_{\tau}^2 d\tau]$.

Here is my reasoning:

$E[\int_{t-1}^t \sigma_{\tau}^2 d\tau] = \int_{t-1}^t E[\sigma_{\tau}^2]d\tau $

If I integrate directly the SDE between 0 and $\tau$, I get: $$ \sigma_\tau^2 = \sigma_0^2 + \int_0^\tau \kappa (\theta-\sigma_t^2)dt + \int_0^\tau \psi \sigma_t^2 dW_t $$ Taking the expectation, I get: $$ E[\sigma_\tau^2] = \sigma_0^2 + \kappa \theta \tau - \kappa \int_0^\tau E[\sigma_t^2] dt $$

I was hoping here to obtain a simplified expression for $E[\sigma_\tau^2]$ and then inject into the integral. However that did not simplify.

Does anyone know how can I get an expression for $IV_t$?

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