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.


2 Answers 2


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 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.

  • $\begingroup$ Ah yes, for CIR models you could do this. +1. $\endgroup$
    – user34971
    Jul 26, 2021 at 12:10
  • 1
    $\begingroup$ @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. $\endgroup$
    – Kevin
    Jul 26, 2021 at 12:22

A partial but general answer:

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.

I do not think there is a simpler way to calculate the variance exactly and in full generality.


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