In Cochrane's "Fiscal Theory of the Price Level", I am struggling with the following derivation. Take the first line as given, where $\pi_t$ and $i_t$ are random processes adapted to $\mathcal{F}_t$; the second line supposedly follows. $L$ is the lag operator, so $L x_t = x_{t-1}$ and $L^{-1}x_t = x_{t+1}$ for any random process $x_t$.

The problematic implication is the following:

$$E_t \Big[ (1- \lambda_1^{-1} L)(1 - \lambda_2 L^{-1}) \pi_{t+1} \Big] = \sigma \kappa \lambda_1^{-1} i_t \\ $$

$$ \Rightarrow \pi_{t+1} = E_{t+1} \frac{\lambda_1^{-1}}{(1- \lambda_1^{-1} L)(1 - \lambda_2 L^{-1})} \sigma \kappa i_t + \frac{1}{(1-\lambda_1^{-1}L)} \delta_{t+1},$$

where $\delta_{t+1}$ is a random process adapted to $\mathcal{F}_{t+1}$ with $E_t[\delta_{t+1}] = 0$ for all $t$.

What mathematical operation is performed to arrive at the second line? I suspect the way the derivation works is that one can indeed apply lag polynomials involving only forward lags "through the $E_t$" because of the law of iterated expectation $E_t[E_{t+1}] = E_t$. I believe I can show this, and given this result I can derive the intermediate step $$ E_t \Big[ (1 - \lambda_1^{-1} L) \pi_{t+1} \Big] = E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1} \sigma \kappa i_t \Big].$$ I can attempt to derive the rest directly, but I arrive at a different result. Specifically, I get

$$ E_t \pi_{t+1} = E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1} \sigma \kappa i_t \Big] + \lambda_1^{-1} \pi_t $$

$$ \Rightarrow \pi_{t+1} = E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1} \sigma \kappa i_t \Big] + \lambda_1^{-1} \pi_t + \delta_{t+1},$$

where $\delta_{t+1}$ is just the expectation error $\pi_{t+1} - E_t\pi_{t+1}$ and so has the required properties. Iteratively substituting for $\pi_t$ backwards using this equation yields

$$ \pi_{t+1} = (1 - \lambda_1^{-1} L)^{-1} E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1} \sigma \kappa i_t \Big] + (1 - \lambda_1^{-1} L)^{-1} \delta_{t+1}.$$

So, the only thing I don't understand is how to move the lag polynomial inside the expectation in the first term on RHS, and also where the $E_{t+1}$ comes from.


In light of John Cochrane's response below (Thank you!), what I still struggle with is the following--maybe someone could shed some light, it would be very appreciated. We have

$$ -\gamma = (1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1} - E_{t+1} \Big[(1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1} \Big]. $$

This means that $\gamma$ is defined as negative one times the expectation error of $(1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1}$ relative to its expectation at time $t+1$. We can expand the lag polynomials, \begin{align} (1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1})\pi_{t+1} &= \Big( 1 + \lambda_1^{-1} \lambda _2 - \lambda_1^{-1} L - \lambda_2 L^{-1} \Big) \pi_{t+1} \\ &= (1 + \lambda_1^{-1} \lambda_2) \pi_{t+1} - \lambda_1^{-1} \pi_t - \lambda_2 \pi_{t+2} \\ &\equiv \xi_{t+2}, \end{align} where I've put a $t+2$ subscript on $\xi$ since it is in the $\mathcal{F}_{t+2}$ information set and not in $\mathcal{F}_{t+1}$, as it is a function only of random variables which themselves are in this information set and at least one of these variables ($\pi_{t+2}$) is not in $\mathcal{F}_{t+1}$. Substituting into the equation for $\gamma$, we get

\begin{align} -\gamma &= \xi_{t+2} - E_{t+1} \xi_{t+2}. \end{align} For the same reason as for $\xi_{t+2}$, it seems to me that $\gamma$ must also be in the $\mathcal{F}_{t+2}$ information set and not in the $\mathcal{F}_{t+1}$ information set.

Therefore, \begin{align} E_{t+1} \frac{\gamma_{t+2}}{(1-\lambda_1^{-1}L)(1-\lambda_2 L^{-1})} \ne 0 \end{align} in general. The terms involving $E_{t+1} \gamma_{t+2}$ are zero, as are the terms involving $E_{t+1} \gamma_{t+2+j} = E_{t+1} \big[ E_{t+1+j} \gamma_{t+2+j} \big] = 0$ for $j > 0$. But there are also terms with $\gamma_{t+2-j}$ for $j > 0$, which are in the $\mathcal{F}_{t+2-j}$ information set and hence are known at time $t+1$ and therefore their expectation at time $t+1$ is not equal to zero (i.e., not equal to the function that maps all $\omega \in \Omega$ to zero, where $\Omega$ is the underlying state space).

I don't think this would change the overall point that inflation $\pi_t$ is a weighted average of forward and past nominal interest rates $i_t$ plus some weighted average of past expectation errors. It would just add another weighted average of a different set of past expectation errors.

Finally, another angle on this is \begin{align} (1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1} - E_{t+1} \Big[(1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1} \Big] \ne (1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t'+1} - E_{t'+1} \Big[(1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t'+1} \Big] \end{align} for $t' \ne t$.

  • $\begingroup$ What is FTPL textbook? My first guess would be Cochrane's "Fiscal Theory of the Price Level", but that one is not a textbook (or is it?). $\endgroup$ Jun 9, 2023 at 13:09
  • $\begingroup$ Yes, that's the one. Idk, I think of it as a textbook, but I guess that's up to interpretation. $\endgroup$
    – Econ
    Jun 10, 2023 at 14:42

2 Answers 2


Here is a revised derivation. Thanks! This is much clearer, I hope.

\begin{equation} E_{t}\left[ (1-\lambda_{1}^{-1}L)(1-\lambda_{2}L^{-1})\pi_{t+1}\right] =\sigma\kappa\lambda_{1}^{-1}i_{t}. \label{inside} \end{equation}

Now we want to invert the lag polynomial. I'll do this a bit slowly. (Thanks to Nicolas Fernandez-Arias for a query.) Operations involving $E_t$ and $L$ are tricky, because you can't push the $L$ inside and outside the $E_t$ carelessly. $E_t(L^{-1}x_{t+1})$ means $E_t(x_{t+2})$. But $L^{-1}[E_t(x_{t+1})]$ means $E_{t+1}(x_{t+2})$. Make sure you do or don't want the $L$ operator to apply to the expectation, not just the variable inside. So, proceeding cautiously, define $\delta_{t+1}$ by \begin{equation} E_{t+1}\left[ (1-\lambda_{1}^{-1}L)(1-\lambda_{2}L^{-1})\pi_{t+1}\right] =\sigma\kappa\lambda_{1}^{-1}i_{t} + \frac{\lambda_1-\lambda_2}{\lambda_1}\delta_{t+1}. \end{equation} The coefficient in front of $\delta_{t+1}$ simplifies the final expression. Note $E_t\delta_{t+1}=0.$ Define $\gamma$ by \begin{equation} (1-\lambda_{1}^{-1}L)(1-\lambda_{2}L^{-1})\pi_{t+1} + \gamma =\sigma\kappa\lambda_{1}^{-1}i_{t} + \frac{\lambda_1-\lambda_2}{\lambda_1}\delta_{t+1}. \end{equation} $\gamma$ is in the ``$t+\infty$'' information set, and $E_{t+1}\gamma=0.$ Now you can invert
\begin{equation} \pi_{t+1} + \frac{\gamma}{(1-\lambda_1^{-1})(1-\lambda_2^{-1})} = \frac{\lambda_1^{-1}}{(1-\lambda_{1}^{-1}L)(1-\lambda_{2}L^{-1})}[\sigma\kappa i_{t}+(\lambda_1-\lambda_2)\delta_{t+1}]. \end{equation} Use the partial fractions decomposition to break up the right hand side, to give \begin{equation} \pi_{t+1}+ \frac{\gamma}{(1-\lambda_1^{-1})(1-\lambda_2^{-1})} =\frac{1}{\lambda_{1}-\lambda_{2}}\left( 1+\frac{\lambda _{1}^{-1}L}{1-\lambda_{1}^{-1}L}+\frac{\lambda_{2}L^{-1}}{1-\lambda_{2}L^{-1}% }\right) [\sigma\kappa i_{t}+ (\lambda_1-\lambda_2)\delta_{t+1}]. \end{equation} Here I follow the usual practice and I rule out solutions that explode in the forward direction. Now take $E_{t+1}$ of both sides. $E_{t+1}\delta_{t+j}=0$, so the $\lambda_2 $ term operating on $\delta_{t+1}$ is zero and \begin{equation} \pi_{t+1}=\frac{1}{\lambda_{1}-\lambda_{2}}E_{t+1}\left( 1+\frac{\lambda _{1}^{-1}L}{1-\lambda_{1}^{-1}L}+\frac{\lambda_{2}L^{-1}}{1-\lambda_{2}L^{-1}% }\right) \sigma\kappa i_{t}+\frac{1}{(1-\lambda_{1}^{-1}L)}\delta_{t+1}% \end{equation} or in sum notation, \begin{equation} \pi_{t+1}=\sigma\kappa\frac{1}{\lambda_{1}-\lambda_{2}}\left( i_{t}% +\sum_{j=1}^{\infty}\lambda_{1}^{-j}i_{t-j}+\sum_{j=1}^{\infty}\lambda_{2}% ^{j}E_{t+1}i_{t+j}\right) +\sum_{j=0}^{\infty}\lambda_{1}^{-j}\delta_{t+1-j}. \label{eq:pi_solution_appendix}% \end{equation}

John Cochrane


I'll assume throughout that $|\lambda_1| > 1$ and $|\lambda_2| < 1$, since otherwise the sums would diverge.

Start from

$$ \mathrm{E}_t \left[(1 - \lambda_1^{-1} L)(1 - \lambda_2L^{-1})\pi_{t + 1}\right] = \sigma\kappa\lambda_1^{-1}i_t $$ This is equivalent to

$$ (1 - \lambda_1^{-1} L)(1 - \lambda_2L^{-1})\pi_{t + 1} = \sigma\kappa\lambda_1^{-1}i_t + \delta_{t + 1}, $$ where $\delta_{t+1}$ is as you defined it in your question. Now operate on both sides by $(1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}$ to get

$$ \pi_{t + 1} = (1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\sigma\kappa \lambda_1^{-1}i_t + (1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1} $$

Now, note that the inverse operators commute (why?). Then, the second term on the right hand side above is

\begin{align} (1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1} &= (1 - \lambda_1^{-1} L)^{-1}(1 - \lambda_2L^{-1})^{-1}\delta_{t + 1} \\ &= (1 - \lambda_1^{-1} L)^{-1} \sum_{j = 0}^\infty \lambda_2^jL^{-j}\delta_{t + 1}. \end{align} Operating on both sides by $\mathrm{E}_{t+1}$ and recalling that $\delta_{t +1}$ is adapted to $\mathcal{F}_{t + 1}$ and mean zero, we have \begin{align} \mathrm{E}_{t+1}\left[(1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1}\right] & = \mathrm{E}_{t + 1}\left[(1 - \lambda_1^{-1} L)^{-1} \sum_{j = 0}^\infty \lambda_2^jL^{-j}\delta_{t + 1}\right] \\ &= (1 - \lambda_1^{-1} L)^{-1}\mathrm{E}_{t + 1}\left[ \sum_{j = 0}^\infty \lambda_2^jL^{-j}\delta_{t + 1}\right] \\ &= (1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1}. \end{align}

Using this, we can operate on both sides of the expression for $\pi_{t + 1}$ above with $\mathrm{E}_{t + 1}$ (noting also that $\pi_{t+1}$ is $\mathcal{F}_{t + 1}$-measurable) to get the expression you are looking for.

Hope this helps!

  • $\begingroup$ Thank you for the detailed explanation. My question really should have mentioned one more thing. Your first step is how I had "hoped" it could be derived initially, but what I don't understand is... the random process on the LHS of the second equation is not adapted to $\mathcal{F}_t$, right? Since it involves forward lags. So then how can I define $\delta_{t+1}$ as its expectation error? That's why I thought one had to first get rid of the forward lags on the LHS before defining $\delta_{t+1}$. What am I missing? Thank you!! $\endgroup$
    – Econ
    Jun 10, 2023 at 14:46
  • $\begingroup$ Sorry should have written "is not adapted to $\mathcal{F}_{t+1}$", since it involves an infinite series of forward lags of $\pi_{t+1}$. $\endgroup$
    – Econ
    Jun 10, 2023 at 14:55
  • $\begingroup$ @Econ The LHS doesn't involve an infinite series of forward lags (note it's not the inverse operator), but I think you are right that it is not $\mathcal{F}_{t + 1}$-measurable, given how you've defined the random variables. That makes my derivation invalid. I'll try to see if there's another way. $\endgroup$ Jun 11, 2023 at 9:19
  • $\begingroup$ Woops yeah my mistake but point still stands and thank you. $\endgroup$
    – Econ
    Jun 11, 2023 at 19:45

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