There are a few online resources available to help with log-linearization (e.g., here or here). However, log-linearization where an expectation is involved is a little tricky because the log can't simply "pass through" the expectation operator. Could somebody help with the algebra in this example?

I have the Euler equation (equation 1) $$ 1 = E_t \left [ \left \{ \delta \left (\frac{C_{t+1}}{C_t} \right )^{-1/\psi} \right \}^\theta \left \{ \frac{1}{1 + R_{m,t+1}} \right \}^{1 - \theta} 1 + R_{i, t+1} \right ] $$ where $\theta = ( 1 -\gamma)/(1 - 1/\psi)$. I'm trying to derive an expression for the risk-free rate and an expression for the equity premium. How should I go about doing this?

It seems from the second link above that I should start by replacing the variables of interest like so $C_t = c e^{\tilde C_t}$. Then following the steps given, it seems like I should arrive at (equation 2)

\begin{align} 1 = E_t \left [ \left \{ \delta \left (\frac{\tilde C_{t+1} + 1}{\tilde C_t + 1} \right )^{-1/\psi} \right \}^\theta {} \left \{ \frac{1}{(1 + R_m)[\widetilde{(1 + R_{m,t+1})} + 1]} \right \}^{1 - \theta} \cdot \\ \cdot [(1 + R_i)[\widetilde{(1 + R_{i, t+1})} + 1]] \right ]. \end{align}

But where do I go from here?


  1. I have copied equation 1 directly from the notes that I have. It is probably the case that the term on the right, $1 + R_{i,t+1}$, should be in parentheses, $(1 + R_{i,t+1})$. In my initial attempt at log-linearization I have treated it this way.

  2. In equation 2, I have followed the steps in the instruction that can be found in the second link at the beginning. So, $R_i$ and $R_m$ without time subscripts are these values in the steady state.

  3. $R_m$ is the return on the market portfolio and $R_i$ is the return on asset $i$.


Thanks for the useful comments. So, from what I have gathered so far, I should get something like this:

\begin{align} 1 &= E_t \left [ \delta^\theta (1 - \frac \theta \psi (\tilde C_{t+1} - \tilde C_t ) (1 + R_m)^{\theta - 1} (\theta - 1) \left ( 1 + \tilde R_{m,t} \frac{R_m}{1 + R_m} \right ) \right .\\ & \left . \, \cdot (1 + R_i) \left ( (1 + \tilde R_{i,t} \frac{ R_i}{1 + R_i} \right ) \right ] \end{align}

Then this would imply that the risk-free rate is found as follows:

\begin{align} 1 &= E_t \left [ \delta^\theta (1 - \frac \theta \psi (\tilde C_{t+1} - \tilde C_t ) (1 + R_m)^{\theta - 1} (\theta - 1) \left ( 1 + \tilde R_{m,t} \frac{R_m}{1 + R_m} \right ) (1 + R_f) \right ] \\ 1 &= E_t[ m_{t+1} (1 + R_f)] \\ \frac{1}{E_t[m_{t+1}]} &= 1 + R_f. \end{align}

Is this correct? And now, to finish the question, how would I find the equity premium?

  • $\begingroup$ Im on the run, but do you have access to Gali's book? I think he does it extensively, iirc $\endgroup$
    – FooBar
    Dec 12, 2014 at 1:52
  • $\begingroup$ No. Is it his Monetary Policy book that it would be in? "Monetary Policy, Inflation, and the Business Cycle?" $\endgroup$
    – ethan1410
    Dec 12, 2014 at 1:58
  • $\begingroup$ The last equality you've given (1 over the risk free rate is equal to the expectation of the sdf) is always true, so that's a good sign. To find the equity premium, find the price for $E_t[m_{t+1}(1+R_m)]$, the value of a claim to the market, then subtract the price of the risk free return: 1. $\endgroup$
    – jayk
    Dec 13, 2014 at 15:36

3 Answers 3


Let's ignore for the moment the existence of the expected value. If this was a deterministic set-up, linearization through taking logs would be straightforward, and without the tricks of the links the OP provided. Taking natural logs on both sides of the first equation we obtain:

$$0 = \theta \ln \delta-\frac {\theta}{\psi}\ln \left (\frac{C_{t+1}}{C_t} \right )-(1-\theta) \ln(1 + R_{m,t+1}) + \ln (1 + R_{i, t+1}) \tag{1}$$


$$\hat c_{t+1} = \frac{C_{t+1}-C_t}{C_t} \Rightarrow \frac{C_{t+1}}{C_t} = 1+\hat c_{t+1} \tag{2}$$

Also, note that it is standard approximation to write $\ln (1+a) \approx a$ at least for $|a|<0.1$. Usually this is the case with growth rates and financial rates so we obtain

$$0 = \theta \ln \delta-\frac {\theta}{\psi}\hat c_{t+1} -(1-\theta) R_{m,t+1} + R_{i, t+1} \tag{3}$$

which is a clear dynamic relation that links the three variables present. If in the model, the steady-state is characterized by constant consumption and constant returns, then at it we will have $\hat c_{t+1} =0$ and so the steady-state relation will be

$$R_{i} = - \theta \ln \delta + (1-\theta) R_{m} \tag{4}$$

But we did all these ignoring the expected value. Our expression is $E_t\left[f\left(C_t, C_{t+1},R_{m,t+1},R_{i,t+1} \right)\right]$, not just $f\left(C_t, C_{t+1},R_{m,t+1},R_{i,t+1} \right)$. Enter first-order Taylor expansion of $f()$. We need a center of expansion. Represent the four variables simply by $\mathbf z_{t+1}$ (it doesn't hurt that a variable with $t$-index is present in $\mathbf z_{t+1}$). We choose to expand the function around $E_t(\mathbf z_{t+1})$. So

$$f\left(\mathbf z_{t+1}\right) \approx f\left(E_t[\mathbf z_{t+1}]\right) + \nabla f\left(E_t[\mathbf z_{t+1}]\right)\cdot \big(\mathbf z_{t+1}-E_t[\mathbf z_{t+1}]\big) \tag{5}$$


$$E_t\left[f\left(\mathbf z_{t+1}\right)\right] \approx f\left(E_t[\mathbf z_{t+1}]\right) \tag{6}$$

Obviously this is an approximation, i.e. it has error, even if only because of Jensen's inequality. But it is standard practice. Then we see that all the previous work we did on the deterministic version, can be applied in the stochastic version inserting conditional expected values in place of the variables. So eq. $(3)$ is written

$$0 = \theta \ln \delta-\frac {\theta}{\psi}E_t[\hat c_{t+1}] -(1-\theta) E_t[R_{m,t+1}] + E_t[R_{i, t+1}] \tag{7}$$

But where are the steady-state values? Well, steady state values in a stochastic context are a bit tricky -are we arguing that our variables (which are now treated as random variables) become constants? Or is there another way to define a steady-state in a stochastic context?

There are more than one ways. One of them, is the "perfect foresight steady state", where we forecast perfectly a not-necessarily constant value (this is the concept of "equilibrium as fulfilled expectations"). This is for example used in Jordi Gali's book mentioned in a comment. "Perfect-foresight steady state" is defined by $$E_t(x_{t+1}) = x_{t+1} \tag{8}$$

Under this concept, eq. $(7)$ becomes eq. $(3)$ which is now the "perfect-foresight stochastic steady state" equation of the economy.

If we want a stronger condition, saying that variables become constant in the steady-state, then it is also reasonable to argue that, again, their forecast will eventually be perfect. In that case, the steady-state of the stochastic economy is the same as that of the deterministic economy, i.e. eq. $(4)$.

  • $\begingroup$ @jmbejara This is perfectly correct. It is the expected value of the truncated first order Taylor approximation of a function. Do you disagree with that? Whether you consider it to be a suboptimal approximation, is another matter, and has to do with what criteria you use to judge the quality and the adequacy of the approximation. $\endgroup$ Dec 12, 2014 at 13:24
  • $\begingroup$ Ok. You have a point. But, like you say, I'm not sure what the best thing in the situation is. But there certainly seems to be different ways of going about it. There is definitely something to be said about the bias, but you bring up a good point. I will undo the vote as soon as it will let me. $\endgroup$
    – jmbejara
    Dec 12, 2014 at 18:46

The correct approximation is $f(x) \approx E[f(x)] + E[f'(x)] (x - E[x])$. This is unbiased, whereas $f(x) \approx E[f(x)] + f'(E[x]) (x - E[x])$ is not. To see this, project $f(x) - \overline{f(x)}$ on $x - \bar x$, where the "bar" represents the expectation operator. Then, approximate $$ \frac{\text{Cov}(f(x), x)}{\text{Var(x)}} \approx E[f'(x)]. $$ This approximation is exact when $x$ is normally distributed (by Stein's lemma).


For clarification, see that the projection of $f(x) - \overline{f(x)}$ on $x - \bar x$ gives us $f(x) - \overline{f(x)} = \beta (x - \bar x) + \epsilon$, where $E[\epsilon] = E[\epsilon x] = 0$ and $\beta = \frac{\text{Cov}(f(x), x)}{\text{Var(x)}}$. If we use Stein's lemma to approximate $\beta$ as described above, we are left with $$ f(x) \approx E[f(x)] + E[f'(x)] (x - \bar x) + \epsilon, $$ which is unbiased, $$ E[\epsilon] = 0. $$ On the other hand, $$ E[f(E[x]) + f'(E[X]) (x - E[x])] = f(E[x]) \neq E[f(x)]. $$

  • $\begingroup$ It would be helpful if you could include in your answer the detailed derivation of the approximation $f(x) \approx E[f(x)] + E[f'(x)] (x - E[x])$. $\endgroup$ Dec 12, 2014 at 14:18
  • $\begingroup$ Thanks for enhancing your answer. To stay close to the question, the OP has a function $f(x)$ and he wants to manipulate its expected value. So he should he solve the expression you write for $E[f(x)]$ and obtain $$E[f(x)] \approx f(x) - \frac {\text{Cov}(f(x),x)}{\text{Var}(x)}\cdot [x-E(x)]?$$ $\endgroup$ Dec 12, 2014 at 19:21

Your problem seems like asset-pricing equation with recursive (Epstein-Zin) preferences. When interested in asset prices, one has to be careful with the usual "macroeconomic" linearization. Such an approximation is certainty-equivalent, meaning that coefficients of linearized solution do not depend on size of shocks. Moreover, all variables in linearized solution will fluctuate around their deterministic steady states. As a result, risk premia are zero, which kind of defies the point.

One solution is to use higher-order perturbation methods (2nd order to get constant risk premia, 3rd order for time-varying premia). This is easy to do with existing software (e.g. Dynare) if you want to solve the model numerically anyway (in which case there is also no need to linearize manually). If instead analytical (approximate) solution is preferred, the usual way is to linearize dynamics of quantities (e.g. consumption growth), then obtain asset prices directly from Euler equation, computing expectations using lognormality assumption, as in Bansal & Yaron (2004).

For example, if lowercase variables are logs, the usual Euler equation can be rewritten as

$$ 1 = E_t [\exp(m_{t+1} + r_{t+1})] $$

If $m_{t+1},r_{t+1}$ are (conditionally) jointly normal, the above implies

$$ 0 = E_t[m_{t+1}] + E_t[r_{t+1}] + \frac{1}{2} \left\{ Var_t[m_{t+1}] + Var_t[r_{t+1}] + 2 Cov_t[m_{t+1},r_{t+1}] \right\} \tag{1} $$

Risk free rate must satisfy $\exp(-r^f_t) = E_t[\exp(m_{t+1})]$, or

$$ r^f_t = -E_t[m_{t+1}] - \frac{1}{2} Var_t[m_{t+1}] $$

and thus we must have

$$ E_t[r_{t+1}] - r^f_t + \frac{1}{2}Var_t[r_{t+1}] = Cov_t[m_{t+1},r_{t+1}] $$

To actually compute asset prices, one would then

  • express log-SDF as a linear function of some state variables and shocks (e.g. log consumption growth in CRRA case)

  • linearize return in terms of log dividend-price ratio (Campbell-Shiller approximation), substitute that into (1).

  • express log D/P ratio as linear in state variables, then use method of undetermined coefficients to obtain a solution for it which satisfies (1).

In practice it's a bit more complicated (especially with EZ preferences, when one has to use the approach first to derive market return that enters SDF, then second time for other returns), but more details can be found e.g. in the linked Bansal & Yaron paper.

  • 1
    $\begingroup$ Exactly. It looks like the confusion in this thread came from the fact that in a first-order approximation of an Euler equation for asset pricing, there is no risk premium. (Covariance between the SDF and return, of course, is inherently second order.) Thanks for clearing this up. $\endgroup$ Dec 14, 2014 at 18:15

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