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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 $E_t[e^{\sigma^2_{r,t+1}}]$?

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  • $\begingroup$ I think you are missing a $]$ somewhere around your $\log$ functions. $\endgroup$
    – SRKX
    Commented Dec 22, 2015 at 6:28
  • $\begingroup$ Is your question regarding how to compute $E_t[e^{\sigma^2_{r,t+1}}]$? $\endgroup$
    – Gordon
    Commented Dec 25, 2015 at 12:58
  • $\begingroup$ Vandalizing questions isn’t allowed. $\endgroup$
    – Bob Jansen
    Commented Oct 31, 2017 at 5:18

2 Answers 2

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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*}

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  • $\begingroup$ 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
    Commented Dec 27, 2015 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
    Commented Dec 27, 2015 at 13:51
  • $\begingroup$ @volcompt: I added one possible interpretation of the log-linear approximation. $\endgroup$
    – Gordon
    Commented Dec 27, 2015 at 17:09
  • $\begingroup$ You deserved it ;) $\endgroup$
    – phdstudent
    Commented Dec 27, 2015 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
    Commented Jan 12, 2016 at 16:05
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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.

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  • $\begingroup$ 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
    Commented Dec 22, 2015 at 9:34

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