# Heston: Variance of Integrated Variance

Consider the standard Heston model\begin{align*} dX&=\left(r-\frac{1}{2}v\right)dt+\sqrt{v}dB,\\ dv&=\kappa(\theta-v)dt+\xi\sqrt{v}dW, \\ dBdW&=\rho dt. \end{align*} Computing $$\mathbb{E}\int_0^t v_sds$$ is simple but does anyone have a reference for \begin{align} Var\left(\int_0^t v_sds\right) \end{align} or is there a simple trick to solve this integral and compute its second moment?

Using Ito's Lemma ($$d(tv)=vdt+tdv$$) and using the SDE for $$dv$$, I only got \begin{align} \int_0^t v_udu=tv_t-\frac{1}{2}\kappa\theta t^2+\kappa\int_0^t u v_udu-\xi\int_0^t u\sqrt{v_u}dW_u, \end{align} which doesn't look too helpful.

Studying zero-coupon bond prices in the CIR (1985) short rate model, $$\text{d}r_t=\kappa(\theta-r_t)\text{d}t+\xi\sqrt{r_t}\text{d}W_t$$, Hirsa (2013, Section 1.2.6.2) states that the characteristic function of the realised interest rate $$R_t=\int_0^t r_s\text{d}s$$ is \begin{align*} \varphi_{R_t}(u)=\mathbb{E}\left[e^{iuR_t}\right] = A_t(u)e^{B_t(u)r_0}, \end{align*} where \begin{align*} A_t(u) &= \frac{\exp\left(\frac{\kappa^2\theta t}{\xi^2}\right)}{\left(\cosh\left(\frac{1}{2}\gamma t\right)+\frac{\kappa}{\gamma}\sinh\left(\frac{1}{2}\gamma t\right)\right)^{2\kappa\theta/\xi^2}}, \\ B_t(u) &= \frac{2iu}{\kappa+\gamma\coth\left(\frac{1}{2}\gamma t\right)},\\ \gamma &= \sqrt{\kappa^2-2\xi^2iu}. \end{align*}

As you say, $$\mathbb{E}[R_t]$$ can be easily computed using Fubini's theorem but from here you also have $$\mathbb{E}[R_t]=-i\varphi_{R_t}'(0).$$ The variance is \begin{align} \mathbb{V}\text{ar}[R_t] &= \mathbb{E}[R_t^2] - \mathbb{E}[R_t]^2 \\ &=-\varphi_{R_t}''(0) + \varphi_{R_t}'(0)^2. \end{align} Computing these derivatives may be ugly. You could use finite differences instead, $$\mathbb{E}[R_t^2]\approx-\frac{\varphi_{R_t}(-h)-2\varphi_{R_t}(0)+\varphi_{R_t}(h)}{h^2}=\frac{2-\varphi_{R_t}(-h)-\varphi_{R_t}(h)}{h^2}.$$

Note A similar term to $$\gamma$$ appears in the characteristic function of the log stock price of the Heston (1993) model. One needs to be careful with the sign of the root (little Heston trap''). I'm not sure whether the same applies here.

• Ah yes, for CIR models you could do this. +1.
– user34971
Jul 26, 2021 at 12:10
• @FridoRolloos Thank you. I already upvoted your answer for mentioning the Clark–Ocone theorem, which I did not know. Looks very interesting. Would be cool to see whether it allows one to compute the variance in closed-form. Jul 26, 2021 at 12:22

Let $$\mathcal{F}_t^W$$ be the filtration generated by $$W$$. Since $$X_T = \int_t^T v_u du$$ is $$\mathcal{F}_T^W$$ measurable, the Clark-Ocone-Haussman formula states $$X_T = E_t[X_T] + \int_t^T E_u \left[ D_u^W X_T \right] dW_u$$ with $$D_u^W X_T$$ denoting the Malliavin derivative of $$X_T$$ with respect to $$W_u$$.
Hence, $$Var(X_T) = E_t \left[ \left( X_T - E_t[X_T] \right)^2 \right] = E_t \left[ \int_t^T \left( E_u \left( D_u^W X_T \right) \right)^2 du \right]$$
I think the Malliavin derivative $$D_u^W X_T$$ can be calculated explicitly for the Heston model, and perhaps then also the expectation of integral on the right hand side. If I have time I'll check this for you and for myself.