# Solving stochastic difference equation in New Keynesian model (FTPL textbook derivation)

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.

UPDATE:

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 $$t' \ne t$$.

• 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?). Commented Jun 9, 2023 at 13:09
• Yes, that's the one. Idk, I think of it as a textbook, but I guess that's up to interpretation.
– Econ
Commented Jun 10, 2023 at 14:42

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

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

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 $$$$E_{t+1}\left[ (1-\lambda_{1}^{-1}L)(1-\lambda_{2}L^{-1})\pi_{t+1}\right] =\sigma\kappa\lambda_{1}^{-1}i_{t} + \frac{\lambda_1-\lambda_2}{\lambda_1}\delta_{t+1}.$$$$ The coefficient in front of $$\delta_{t+1}$$ simplifies the final expression. Note $$E_t\delta_{t+1}=0.$$ Define $$\gamma$$ by $$$$(1-\lambda_{1}^{-1}L)(1-\lambda_{2}L^{-1})\pi_{t+1} + \gamma =\sigma\kappa\lambda_{1}^{-1}i_{t} + \frac{\lambda_1-\lambda_2}{\lambda_1}\delta_{t+1}.$$$$ $$\gamma$$ is in the $$t+\infty$$'' information set, and $$E_{t+1}\gamma=0.$$ Now you can invert
$$$$\pi_{t+1} + \frac{\gamma}{(1-\lambda_1^{-1})(1-\lambda_2^{-1})} = \frac{\lambda_1^{-1}}{(1-\lambda_{1}^{-1}L)(1-\lambda_{2}L^{-1})}[\sigma\kappa i_{t}+(\lambda_1-\lambda_2)\delta_{t+1}].$$$$ Use the partial fractions decomposition to break up the right hand side, to give $$$$\pi_{t+1}+ \frac{\gamma}{(1-\lambda_1^{-1})(1-\lambda_2^{-1})} =\frac{1}{\lambda_{1}-\lambda_{2}}\left( 1+\frac{\lambda _{1}^{-1}L}{1-\lambda_{1}^{-1}L}+\frac{\lambda_{2}L^{-1}}{1-\lambda_{2}L^{-1}% }\right) [\sigma\kappa i_{t}+ (\lambda_1-\lambda_2)\delta_{t+1}].$$$$ Here I follow the usual practice and I rule out solutions that explode in the forward direction. Now take $$E_{t+1}$$ of both sides. $$E_{t+1}\delta_{t+j}=0$$, so the $$\lambda_2$$ term operating on $$\delta_{t+1}$$ is zero and $$$$\pi_{t+1}=\frac{1}{\lambda_{1}-\lambda_{2}}E_{t+1}\left( 1+\frac{\lambda _{1}^{-1}L}{1-\lambda_{1}^{-1}L}+\frac{\lambda_{2}L^{-1}}{1-\lambda_{2}L^{-1}% }\right) \sigma\kappa i_{t}+\frac{1}{(1-\lambda_{1}^{-1}L)}\delta_{t+1}%$$$$ or in sum notation, $$$$\pi_{t+1}=\sigma\kappa\frac{1}{\lambda_{1}-\lambda_{2}}\left( i_{t}% +\sum_{j=1}^{\infty}\lambda_{1}^{-j}i_{t-j}+\sum_{j=1}^{\infty}\lambda_{2}% ^{j}E_{t+1}i_{t+j}\right) +\sum_{j=0}^{\infty}\lambda_{1}^{-j}\delta_{t+1-j}. \label{eq:pi_solution_appendix}%$$$$

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!

• 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!!
– Econ
Commented Jun 10, 2023 at 14:46
• Sorry should have written "is not adapted to $\mathcal{F}_{t+1}$", since it involves an infinite series of forward lags of $\pi_{t+1}$.
– Econ
Commented Jun 10, 2023 at 14:55
• @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. Commented Jun 11, 2023 at 9:19
• Woops yeah my mistake but point still stands and thank you.
– Econ
Commented Jun 11, 2023 at 19:45