On the paper Bollerslev, Tauchen and Zhou (2009 RFS) the authors say about equation (15):

The corresponding model implied risk-neutral conditional expectation $$E^Q_t(\sigma^2_{r,t+1})=E_t(\sigma^2_{r,t+1}M_{t+1})E_t(M_{t+1})^{-1}$$ cannot easily be computed in a closed form.

However it is possible to calculate the following close log-linear approximation: $$E^Q_t(\sigma^2_{r,t+1}) \approx \log[e^{-r_{f,t}} E_t[e^{m_{t+1}+\sigma^2_{r,t+1}}]] -\frac{1}{2}Var_t(\sigma_{r,t+1}^2) = E_t(\sigma^2_{r,t+1})+(\theta - 1)\kappa_1 [A_\sigma + A_q \kappa_1^2(A_\sigma^2 + A_q^2 \varphi_q^2)\varphi_q^2]q_t$$

I perfectly understand how to get from the first equality to the second. But the last equality, I have no idea where it comes from.

First, I imagine that the terms: $\log[e^{-r_{f,t}} E_t[e^{m_{t+1}}]]$ cancel out. But then how does he get rid of the $E_t[e^{\sigma^2_{r,t+1}}]$?

  • $\begingroup$ I think you are missing a $]$ somewhere around your $\log$ functions. $\endgroup$ – SRKX Dec 22 '15 at 6:28
  • $\begingroup$ Is your question regarding how to compute $E_t[e^{\sigma^2_{r,t+1}}]$? $\endgroup$ – Gordon Dec 25 '15 at 12:58
  • $\begingroup$ Vandalizing questions isn’t allowed. $\endgroup$ – Bob Jansen Oct 31 '17 at 5:18

We first list the assumptions. \begin{align*} g_{t+1} &= \mu_g + \sigma_{g, t} z_{g, t+1}, \tag{1}\\ \sigma_{g, t+1}^2 &= a_{\sigma} + \rho_{\sigma} \sigma_{g, t}^2 + \sqrt{q_t} z_{\sigma, t+1}, \tag{2} \\ q_{t+1} &= a_{q} + \rho_q q_t + \varphi_q \sqrt{q_t} z_{q, t+1}. \tag{3} \end{align*} Moreover, \begin{align*} r_{t+1} &= -\ln \delta +\psi^{-1} \mu_g - \frac{(1-\gamma)^2}{2\theta} \sigma_{g, t}^2 + (\kappa_1 \rho_q-1)A_q q_t \\ & \quad +\sigma_{g, t}z_{g, t+1} +\kappa_1\sqrt{q_t} (A_{\sigma}z_{\sigma, t+1} + A_q \varphi_q z_{q, t+1}). \tag{10} %\sigma_{r, t}^2 &= \sigma_{g, t}^2 + \kappa_1^2(A_{\sigma}^2 + A_q^2 \varphi_q^2)q_t, \tag{12} \end{align*} From (2) and (3), \begin{align*} \sigma_{r, t+1}^2 &= \sigma_{g, t+1}^2 + \kappa_1^2(A_{\sigma}^2 + A_q^2 \varphi_q^2)q_{t+1}, \tag{13}\\ &=a_{\sigma} + \rho_{\sigma} \sigma_{g, t}^2 + \sqrt{q_t} z_{\sigma, t+1} \\ &\quad + \kappa_1^2(A_{\sigma}^2 + A_q^2 \varphi_q^2)(a_{q} + \rho_q q_t + \varphi_q \sqrt{q_t} z_{q, t+1}). \end{align*} From (1) and (10), \begin{align*} m_{t+1} &= \theta \ln \delta - \theta \psi^{-1}g_{t+1}+(\theta-1)r_{t+1} \tag{4}\\ &=\theta \ln \delta - \theta \psi^{-1}(\mu_g + \sigma_{g, t} z_{g, t+1})\\ &\quad +(\theta-1)\bigg[-\ln \delta +\psi^{-1} \mu_g - \frac{(1-\gamma)^2}{2\theta} \sigma_{g, t}^2 + (\kappa_1 \rho_q-1)A_q q_t\\ &\quad +\sigma_{g, t}z_{g, t+1} +\kappa_1\sqrt{q_t} (A_{\sigma}z_{\sigma, t+1} + A_q \varphi_q z_{q, t+1})\bigg]. \end{align*} Therefore, \begin{align*} %E_t(m_{t+1}) &= \theta \ln \delta - \theta \psi^{-1}\mu_g + (\theta-1)\bigg[-\ln \delta +\psi^{-1} \mu_g - \frac{(1-\gamma)^2}{2\theta} \sigma_{g, t}^2 + (\kappa_1 \rho_q-1)A_q q_t\bigg],\\ %{\rm Cov}_t(m_{t+1}, r_{t+1}) &= -\gamma \sigma_{g, t}^2 + (\theta -1) \kappa_1^2 q_t\big(A_{\sigma}^2 + A_q^2 \varphi_q^2\big),\tag{11}\\ {\rm Cov}_t(m_{t+1}, \sigma_{r, t+1}^2) &=(\theta -1)\kappa_1 \Big[A_{\sigma}+A_q\kappa_1^2\big(A_{\sigma}^2 + A_q^2 \varphi_q^2\big) \varphi_q^2 \Big]q_t . \end{align*} Furthermore, from the conditional normality of $m_{t+1}$ and $\sigma_{r, t+1}^2$, \begin{align*} E_t^Q\left(\sigma_{r, t+1}^2\right) &=E_t\left(\sigma_{r, t+1}^2M_{t+1}\right)/E_t(M_{t+1})\\ &\approx \ln\left(e^{-r_{f, t}} E_t\left(e^{m_{t+1}+\sigma_{r, t+1}^2} \right) \right) - \frac{1}{2} {\rm Var}_t\left(\sigma_{r, t+1}^2\right) \tag{*}\\ &=\ln\left(e^{-r_{f, t}} e^{E_t(m_{t+1}) + \frac{1}{2}{\rm Var}_t(m_{t+1})+E_t(\sigma_{r, t+1}^2)+ \frac{1}{2} {\rm Var}_t\left(\sigma_{r, t+1}^2\right) + {\rm Cov}_t(m_{t+1}, \sigma_{r, t+1}^2)} \right) \\ &\quad- \frac{1}{2} {\rm Var}_t\left(\sigma_{r, t+1}^2\right)\\ &=\ln\left(e^{-r_{f, t}} E_t\left(e^{m_{t+1}}\right) e^{E_t(\sigma_{r, t+1}^2)+ \frac{1}{2} {\rm Var}_t\left(\sigma_{r, t+1}^2\right) + {\rm Cov}_t(m_{t+1}, \sigma_{r, t+1}^2)} \right) - \frac{1}{2} {\rm Var}_t\left(\sigma_{r, t+1}^2\right)\\ &=E_t\left(\sigma_{r, t+1}^2\right) + {\rm Cov}_t(m_{t+1}, \sigma_{r, t+1}^2) \\ &=E_t\left(\sigma_{r, t+1}^2\right) + (\theta-1)\kappa_1\Big[A_{\sigma} + A_q \kappa_1^2 \big(A_{\sigma}^2 + A_q^2 \varphi_q^2\big)\varphi_q^2 \Big]q_t. \end{align*}

Interpretation of Log-linear approximation (*).

Regarding Log-linear approximation (*), as the paper did not supply an explanation, we provide one possible interpretation below. Specifically, note that \begin{align*} e^{\sigma_{r, t+1}^2} &\approx 1+ \sigma_{r, t+1}^2 + \frac{1}{2} \left(\sigma_{r, t+1}^2\right)^2\\ &\approx 1+ \sigma_{r, t+1}^2 + \frac{1}{2} \Big[\big(\sigma_{r, t+1}^2\big)^2 - \left(E_t\big(\sigma_{r, t+1}^2\big)\right)^2\Big]\\ &\approx 1+ \sigma_{r, t+1}^2 + \frac{1}{2}{\rm Var}_t \big(\sigma_{r, t+1}^2\big). \end{align*} Then, \begin{align*} \ln\left(e^{-r_{f, t}} E_t\left(e^{m_{t+1}+\sigma_{r, t+1}^2} \right) \right) &\approx \ln\left(e^{-r_{f, t}} E_t\left(e^{m_{t+1}}\left(1+ \sigma_{r, t+1}^2 + \frac{1}{2}{\rm Var}_t \big(\sigma_{r, t+1}^2\big) \right) \right) \right)\\ &\approx \ln \left(1 +e^{-r_{f, t}} E_t\left(\sigma_{r, t+1}^2M_{t+1}\right) + \frac{1}{2}{\rm Var}_t \big(\sigma_{r, t+1}^2\big) \right)\\ &\approx e^{-r_{f, t}} E_t\left(\sigma_{r, t+1}^2M_{t+1}\right) + \frac{1}{2}{\rm Var}_t \big(\sigma_{r, t+1}^2\big)\\ &= E_t\left(\sigma_{r, t+1}^2M_{t+1}\right)/E_t(M_{t+1}) + \frac{1}{2}{\rm Var}_t \big(\sigma_{r, t+1}^2\big). \end{align*} That is, \begin{align*} E_t\left(\sigma_{r, t+1}^2M_{t+1}\right)/E_t(M_{t+1}) &\approx \ln\left(e^{-r_{f, t}} E_t\left(e^{m_{t+1}+\sigma_{r, t+1}^2} \right) \right) -\frac{1}{2}{\rm Var}_t \big(\sigma_{r, t+1}^2\big). \end{align*}

  • $\begingroup$ Gordon, thank you very much for your answer. Two quick questions: (1) $E_t(M_{t+1})^{-1} = R_f$, so why the first term of the approximation is $-r_f$ (i.e. why the minus sign)? (2) I understand that lower case variables are logs, so I am puzzled how can one move $\sigma_{r, t+1}^2$ to the exponent? $\endgroup$ – phdstudent Dec 27 '15 at 13:19
  • $\begingroup$ @volcompt: For your question (1), we assume that $e^{r_{f, t}} = E_t(M_{t+1})$. For the second question, I need to think about more. $\endgroup$ – Gordon Dec 27 '15 at 13:51
  • $\begingroup$ @volcompt: I added one possible interpretation of the log-linear approximation. $\endgroup$ – Gordon Dec 27 '15 at 17:09
  • $\begingroup$ You deserved it ;) $\endgroup$ – phdstudent Dec 27 '15 at 18:22
  • $\begingroup$ Gordon, a quick comment: When you say: \begin{align*} e^{\sigma_{r, t+1}^2} &\approx 1+ \sigma_{r, t+1}^2 + \frac{1}{2} \left(\sigma_{r, t+1}^2\right)^2\\ &\approx 1+ \sigma_{r, t+1}^2 + \frac{1}{2} \Big[\big(\sigma_{r, t+1}^2\big)^2 - \left(E_t\big(\sigma_{r, t+1}^2\big)\right)^2\Big]\\ &\approx 1+ \sigma_{r, t+1}^2 + \frac{1}{2}{\rm Var}_t \big(\sigma_{r, t+1}^2\big). \end{align*} Why can you subtract on the second equality the term $ \left(E_t\big(\sigma_{r, t+1}^2\big)\right)^2$, without adding it back again ? $\endgroup$ – phdstudent Jan 12 '16 at 16:05

Directly from the paper:

We assume that the representative agent in the economy is equipped with Epstein–Zin–Weil recursive preferences. Consequently, the logarithm of the intertemporal marginal rate of substitution, $m_{t+1} \equiv log(M_{t+1})$, may be expressed as $$m_{t+1} =\theta log\delta−\theta\psi^{-1}g_{t+1}+(\theta−1)r_{t+1}, (4)$$


Let $w_t$ denote the logarithm of the price–dividend ratio, or equivalently the price–consumption or wealth–consumption ratio, of the asset that pays the consumption endowment, $\{C_{t +i} \}_{i=1}^\infty$ . The standard solution method for finding the equilibrium in a model like the one defined above then consists in conjecturing a solution for $w_t$ as an affine function of the state variables, $σ^2_{g,t}$ and $q_t$ , $$w_t = A_0 + A_σσ^2_{g,t} + A_qq_t, (6)$$ solving for the coefficients $A_0$, $A_σ$, and $A_q$, using the standard Campbell and Shiller (1988) approximation $r_{t+1} = κ_0 + κ_1w_{t+1} − w_t + g_{t+1}$.


From the solution for the A’s, it is now relatively straightforward to deduce the reduced form expressions for other variables of interest. In particular, the time $t$ to $t + 1$ return must satisfy the following relation: $$r_{t+1} =−log\delta+ \psi^{-1}\mu_g − \frac{(1-\gamma)^2}{2\theta}\sigma^2_{g,t}+(k_1\rho_q-1)A_qq_t+\sigma_{g,t}z_{g,t+1}+(10)$$ $$k_1\sqrt{q_t}[A_qz_{\sigma,t+1}+A_q\varphi_qz_{q,t+1}]$$


To formally establish this result, denote the conditional variance of the time t to t + 1 return as $σ^2_{r,t} \equiv Var_t (r_{t+1} )$. It follows from Equation (10) that $$\sigma^2_{r,t}=\sigma^2_{g,t} + k_1^2(A_\sigma^2+A^2_q\varphi_q^2)q_t , (12)$$ […]Consider instead the one-period ahead conditional variance, $$\sigma^2_{r,t+1}=\sigma^2_{g,t+1} + k_1^2(A_\sigma^2+A^2_q\varphi_q^2)q_{t+1} , (13)$$ which is unknown or stochastic at time t.[…] It follows readily that the time t objective conditional expectation equals $$E_t[\sigma^2_{r,t+1}]=a_\sigma + k^2_1(A_\sigma^2+A_q^2\varphi^2_q)a_q+\rho_\sigma\sigma^2_{g,t}+k_1^2(A_\sigma^2+\varphi^2_q)\rho_qq_t , (14)$$

If you use (4) and (14) into the second part of the equality you can recover the third part.

  • $\begingroup$ Thank you for your answer @franic. I am very sorry, but I am still not getting how using (4) and (14) I get the second part. On (4) I have: $E^Q_t(\sigma^2_{r,t+1})$, whereas on the equality I have $E_t(e^{\sigma^2_{r,t+1}})$. Is one of this sigmas a log or something? $\endgroup$ – phdstudent Dec 22 '15 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.