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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?

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  • $\begingroup$ Your expression for realised variance ignores the $J^s$ contribution, is that what your intention? $\endgroup$ – Kiwiakos Dec 7 '15 at 21:17
  • $\begingroup$ With jumps, expected realized variance of the spot process is infinite. I guess you want to exclude the jumps and then account for them separately, having in mind the actual calculation will be daily, not continuous? $\endgroup$ – q.t.f. Dec 9 '15 at 17:51
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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.

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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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