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Expected Stochastic Volatility Under a Martingale Mean Level

Article Quant Q&A · Author: jjj
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Summary

The document derives a conditional expectation for a mean-reverting stochastic volatility process. It applies Itô’s lemma to an exponentially rescaled process, integrates the resulting stochastic differential equation over the forecast horizon, and takes the conditional expectation. The stochastic integral contributes zero under the assumptions used, leaving a deterministic integral involving the expected long-run volatility level.

The derivation obtains the familiar exponential decay of the gap between current volatility and its mean level, provided that the evolving mean level is a martingale over the interval. That condition lets its future conditional expectation be replaced by its current value. The answer presents this calculation for one of the expressions in the source paper and says the other can be handled similarly. Its conclusion therefore depends on the stated martingale assumption and the process specification; it does not establish that the assumption holds generally or fully resolve the original question about unconditional expectations.

Key ideas

  • An exponential transformation simplifies the mean-reverting stochastic volatility equation.
  • Taking conditional expectations removes the stochastic integral when it has zero conditional mean.
  • The expected volatility approaches its mean level at an exponential rate.
  • The closed-form expectation relies on the mean level being a martingale over the forecast interval.
  • The derivation covers one expression and leaves a similar calculation for the other.

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# Calculating Expectation of Stochastic Volatility


# Calculating Expectation of Stochastic Volatility












I have a question while reading THE NELSON–SIEGEL MODEL OF THE TERM STRUCTURE OF OPTION IMPLIED VOLATILITY AND VOLATILITY COMPONENTS by Guo, Han, and Zhao.

I don't understand why the above equations hold. It's easy to show that $E_t[\sigma_{t+j}] = e^{-j\alpha}\sigma_t + \alpha \int_t^{t+j}\alpha e^{\alpha (s-t)} E_t[\bar{\sigma_s}]ds$ How does this imply the expectations above? Do we have to assume that $E_t[\bar{\sigma_s}]=E[\bar{\sigma_s}], \quad \forall t \in [0, \infty)$?

## Answer by mmencke (score 2)

https://quant.stackexchange.com/a/69972

This exercise is quite similar to finding the mean in the Vašíček short rate model. I have shown the steps for the second expression below - something along the same lines can be done for the first expression.

Define function $$f_{t}:=f(t,\bar{\sigma}_{t})=e^{\kappa t}\bar{\sigma}_{t}$$ Use Itô on the function \begin{align*} \text{d} f_{t}&=\kappa e^{\kappa t}\bar{\sigma}_{t}\text{d}t+ e^{\kappa t}\text{d}\bar{\sigma}_{t}\\ &=\kappa e^{\kappa t}\bar{\sigma}_{t}\text{d}t+ e^{\kappa t}\left(-\kappa(\bar{\sigma}_{t}-\bar{\bar{\sigma}}_{t})\text{d}t+\xi \bar{\sigma}_{t}\text{d}w_{t}\right)\\ &=e^{\kappa t}\kappa \bar{\bar{\sigma}}_{t}\text{d}t+e^{\kappa t}\xi \bar{\sigma}_{t}\text{d}w_{t}\\ &=e^{\kappa t}\left(\kappa \bar{\bar{\sigma}}_{t}\text{d}t+\xi \bar{\sigma}_{t}\text{d}w_{t}\right) \end{align*} Substituting \begin{align*} \bar{\sigma}_{t+j}&=e^{-\kappa (t+j)}f_{t+j}\\ &=e^{-\kappa (t+j)}\left(f_{t}+\int_{t}^{t+j}\text{d}f_{s}\right)\\ &=e^{-\kappa (t+j)}e^{\kappa t}\bar{\sigma}_{t}+e^{-\kappa (t+j)}\int_{t}^{t+j}e^{\kappa s}\kappa \bar{\bar{\sigma}}_{s}\text{d}s+e^{-\kappa (t+j)}\int_{t}^{t+j}\xi \bar{\sigma}_{t}\text{d}w_{t}\\ &=e^{-\kappa j}\bar{\sigma}_{t}+e^{-\kappa (t+j)}\int_{t}^{t+j}e^{\kappa s}\kappa \bar{\bar{\sigma}}_{s}\text{d}s+e^{-\kappa (t+j)}\int_{t}^{t+j}\xi \bar{\sigma}_{t}\text{d}w_{t} \end{align*}

We can then find the expectation \begin{align*} \mathbb{E}_{t} \left[ \bar{\sigma}_{t+j} \right] &=\mathbb{E}_{t} \left[e^{-\kappa j}\bar{\sigma}_{t}+e^{-\kappa (t+j)}\int_{t}^{t+j}e^{\kappa s}\kappa \bar{\bar{\sigma}}_{s}\text{d}s+e^{-\kappa (t+j)}\int_{t}^{t+j}\xi \bar{\sigma}_{t}\text{d}w_{t}\right]\\ &=e^{-\kappa j}\bar{\sigma}_{t}+e^{-\kappa (t+j)}\int_{t}^{t+j}e^{\kappa s}\kappa \mathbb{E}_{t} \left[\bar{\bar{\sigma}}_{s}\right]\text{d}s \end{align*} as the stochastic integral has mean 0. Assuming that $\bar{\bar{\sigma}}_{t}$ is a martingale, i.e. that $\bar{\bar{\sigma}}_{t}=\mathbb{E}_{t} \left[ \bar{\bar{\sigma}}_{s} \right]$ for $t<s$: \begin{align*} \mathbb{E}_{t} \left[ \bar{\sigma}_{t+j} \right]&=e^{-\kappa j}\bar{\sigma}_{t}+\bar{\bar{\sigma}}_{t}\kappa e^{-\kappa (t+j)}\int_{t}^{t+j}e^{\kappa s} \text{d}s \end{align*} We know that $$\int_{t}^{t+j}e^{\kappa s} \text{d}s=\frac{e^{\kappa (t+j)}-e^{\kappa t}}{\kappa}$$ So \begin{align*} \mathbb{E}_{t} \left[ \bar{\sigma}_{t+j} \right] &=e^{-\kappa j}\bar{\sigma}_{t}+\bar{\bar{\sigma}}_{t}(1-e^{-\kappa j})\\ &=\bar{\bar{\sigma}}_{t}+e^{-\kappa j}\left(\bar{\sigma}_{t}-\bar{\bar{\sigma}}_{t}\right) \end{align*} This was the second expression shown with $\tau:=e^{-\kappa}$.

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