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Expected Realized Variance in a Stochastic Volatility Jump Model

Article Quant Q&A · Author: Limbo
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Summary

The document asks how to calculate expected annualized integrated variance under a stochastic volatility model with jumps in both price and variance. The variance process mean-reverts toward a long-run level, receives diffusion shocks, and can jump; the model is described in a risk-neutral setting and has been calibrated to index options. The question is whether expected realized variance can be identified with the initial variance.

The responses recommend taking expectations inside the time integral and deriving the mean variance from its drift equation. This produces an ordinary differential equation, whose solution can then be integrated over the measurement horizon. A second response uses an integrating factor to express variance as decaying initial variance, mean reversion, a stochastic integral, and jump contributions. Its displayed derivation contains notation and algebra that merit checking, so the simpler expectation-and-ODE route is the clearer guidance. The answer depends on the jump intensity specification and jump-size mean; the document supplies no completed closed-form result.

Key ideas

  • Expected integrated variance is the time integral of expected variance when expectation and integration can be interchanged.
  • The mean variance can be found by taking expectations in the variance drift equation.
  • Mean reversion and expected jump contributions determine an ordinary differential equation for expected variance.
  • The initial variance alone generally does not determine expected realized variance across a horizon.
  • The detailed integrating-factor derivation should be checked carefully for notation and algebra.

Tags

VolatilityOptionsStatisticsDerivatives pricing
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# Realized variance in SVJJ (Heston with jumps) model


# Realized variance in SVJJ (Heston with jumps) model












I am working with the stochastic volatility model with jumps in both the price and volatility dynamics, ie. the risk neutral dynamics are of the form: $$\mathrm{d}V_t = \kappa(\theta - V_t)\mathrm{d}t + \sigma \sqrt{V_t} \mathrm{d}B_t^v + J^v \mathrm{d} N_t \\ \mathrm{d}S_t = (r_t-d_t-\lambda m^j)S_t\mathrm{d}t + \sqrt{V_t} S\mathrm{d}B_t^s + (e^{J^s}-1)S_t \mathrm{d}N_t,$$

where $\text{Corr}(B_t^v,B_t^s) = \rho$, and the jump distributions are $J^v\sim \exp(\mu_v)$ and $J^s \sim \mathcal{N}(\mu_s,\sigma^2)$. I have calibrated the model to a data set with European Call and Put options on SPX, and now I want to derive at a closed form expression of the expected value of the annualized realized variance, that is $\mathbb{E}[1/T \int_0^Tv_t \mathrm{d}t]$.

My first thought was simply that it is equal to the calibrated value of $V_0$ but this is based on a martingale assumption of the integral which I am not sure is correct. How would I compute the expectation if it is not a martingale? If it indeed is a martingale, how would I prove it?

## Answer by Phun (score 2)

https://quant.stackexchange.com/a/22169

Try $$\mathbb{E}\frac{1}{T} \int_0^T V_t dt = \frac{1}{T} \int_0^T \mathbb{E} V_t dt$$ and use $$\frac{1}{dt}\mathbb{E} V_t = \kappa\theta - \kappa \mathbb{E}V_t + (\lambda_0 + \lambda_1\mathbb{E} V_t)\mu_V, $$ which is in fact a simple ODE.

## Answer by user16651 (score 2)

https://quant.stackexchange.com/a/27989

Hint

By application of Ito's lemma, we have $$d(e^{kt}v_t)=\kappa e^{\kappa t}v_t\,dt+e^{\kappa t}dv_t+d(e^{\kappa t})dv_t$$ therefore $$v_t=v_0e^{-\kappa t}+\theta(1-e^{-\kappa t})+\sigma\int_{0}^{t}\sqrt{v_s}e^{-\kappa(t-s)}dB_{s}^{v}+\int_{0}^{t}e^{-\kappa(t-s)}J^v\,dN_{s}$$ $J_v$ is random jump size occurring at time $t_i$ and $N_t=N_t-N_0$ is the total number of jumps in the times interval $(0,t]$, therefore $$v_t=v_0e^{-\kappa t}+\theta(1-e^{-\kappa t})+\sigma\int_{0}^{t}\sqrt{v_t}e^{-\kappa(t-s)}dB_{s}^{v}+\sum\limits_{i=1}^{N_t}e^{-\kappa(t-t_i)}J_{i}^{v}$$ $$\int_{0}^{T}v_tdt=\frac{v_0}{\kappa}\left(1-e^{-\kappa T}\right)+\frac{\theta}{\kappa}\left(-1+\kappa T+e^{-\kappa T}\right)+\frac{1}{\kappa}\left(1-e^{-\kappa T}\right)\sum\limits_{i=1}^{N_t}e^{\kappa t_i}J_{i}^{v}\\ \qquad\,\,\,+\sigma\int_{0}^{T}\int_{0}^{t}\sqrt{v_t}e^{-\kappa(t-s)}dB_{s}^{v} dt$$ and $$\frac{1}{T}\int_{0}^{T}v_tdt=\frac{v_0}{\kappa T}\left(1-e^{-\kappa T}\right)+\frac{\theta}{\kappa T}\left(-1+\kappa T+e^{-\kappa T}\right)+\frac{1}{\kappa T}\left(1-e^{-\kappa T}\right)\sum\limits_{i=1}^{N_t}e^{\kappa t_i}J_{i}^{v}\\ \qquad\,\,\,+\frac{\sigma}{T}\int_{0}^{T}\int_{0}^{t}\sqrt{v_t}e^{-\kappa(t-s)}dB_{s}^{v} dt $$ Now you should calculate $$\mathbb{E}\left[\sum\limits_{i=1}^{N_t}e^{\kappa t_i}J_{i}^{v}\right]$$

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