<p>Here is a revised derivation. Thanks! This is much clearer, I hope.</p>
<p><span class="math-container">\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}</span></p>
<p>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 <span class="math-container">$E_t$</span> and <span class="math-container">$L$</span> are tricky, because you can't push the <span class="math-container">$L$</span> inside and outside the <span class="math-container">$E_t$</span> carelessly. <span class="math-container">$E_t(L^{-1}x_{t+1})$</span> means <span class="math-container">$E_t(x_{t+2})$</span>. But <span class="math-container">$L^{-1}[E_t(x_{t+1})]$</span> means <span class="math-container">$E_{t+1}(x_{t+2})$</span>. Make sure you do or don't want the <span class="math-container">$L$</span> operator to apply to the expectation, not just the variable inside. So, proceeding cautiously, define <span class="math-container">$\delta_{t+1}$</span> by
<span class="math-container">\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}</span>
The coefficient in front of <span class="math-container">$\delta_{t+1}$</span> simplifies the final expression. Note <span class="math-container">$E_t\delta_{t+1}=0.$</span> Define <span class="math-container">$\gamma$</span> by
<span class="math-container">\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}</span>
<span class="math-container">$\gamma$</span> is in the ``<span class="math-container">$t+\infty$</span>'' information set, and <span class="math-container">$E_{t+1}\gamma=0.$</span> Now you can invert<br />
<span class="math-container">\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}</span>
Use the partial fractions decomposition to break up the right hand side, to give
<span class="math-container">\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}</span>
Here I follow the usual practice and I rule out solutions that explode in the forward direction. Now take <span class="math-container">$E_{t+1}$</span> of both sides.  <span class="math-container">$E_{t+1}\delta_{t+j}=0$</span>, so the <span class="math-container">$\lambda_2 $</span> term operating on <span class="math-container">$\delta_{t+1}$</span> is zero and
<span class="math-container">\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}</span>
or in sum notation,
<span class="math-container">\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}</span></p>
<p>John Cochrane</p>

CC BY-SA 4.0
2023-06-15T00:47:47.833Z
2023-06-15T00:47:47.833Z
John Cochrane
<p>I'll assume throughout that <span class="math-container">$|\lambda_1| &gt; 1$</span> and <span class="math-container">$|\lambda_2| &lt; 1$</span>, since otherwise the sums would diverge.</p>
<p>Start from</p>
<p><span class="math-container">$$
\mathrm{E}_t \left[(1 - \lambda_1^{-1} L)(1 - \lambda_2L^{-1})\pi_{t + 1}\right] = \sigma\kappa\lambda_1^{-1}i_t
$$</span>
This is equivalent to</p>
<p><span class="math-container">$$
(1 - \lambda_1^{-1} L)(1 - \lambda_2L^{-1})\pi_{t + 1} = \sigma\kappa\lambda_1^{-1}i_t + \delta_{t + 1},
$$</span>
where <span class="math-container">$\delta_{t+1}$</span> is as you defined it in your question. Now operate on both sides by <span class="math-container">$(1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}$</span> to get</p>
<p><span class="math-container">$$
\pi_{t + 1} = (1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\sigma\kappa \lambda_1^{-1}i_t + (1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1}
$$</span></p>
<p>Now, note that the inverse operators commute (why?). Then, the second term on the right hand side above is</p>
<p><span class="math-container">\begin{align}
(1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1} &amp;= (1 - \lambda_1^{-1} L)^{-1}(1 - \lambda_2L^{-1})^{-1}\delta_{t + 1} \\
&amp;= (1 - \lambda_1^{-1} L)^{-1} \sum_{j = 0}^\infty \lambda_2^jL^{-j}\delta_{t + 1}.
\end{align}</span>
Operating on both sides by <span class="math-container">$\mathrm{E}_{t+1}$</span> and recalling that <span class="math-container">$\delta_{t +1}$</span> is adapted to <span class="math-container">$\mathcal{F}_{t + 1}$</span> and mean zero, we have
<span class="math-container">\begin{align}
\mathrm{E}_{t+1}\left[(1 - \lambda_2L^{-1})^{-1}(1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1}\right] &amp; = \mathrm{E}_{t + 1}\left[(1 - \lambda_1^{-1} L)^{-1} \sum_{j = 0}^\infty \lambda_2^jL^{-j}\delta_{t + 1}\right] \\
&amp;= (1 - \lambda_1^{-1} L)^{-1}\mathrm{E}_{t + 1}\left[ \sum_{j = 0}^\infty \lambda_2^jL^{-j}\delta_{t + 1}\right] \\
&amp;= (1 - \lambda_1^{-1} L)^{-1}\delta_{t + 1}.
\end{align}</span></p>
<p>Using this, we can operate on both sides of the expression for <span class="math-container">$\pi_{t + 1}$</span> above with <span class="math-container">$\mathrm{E}_{t + 1}$</span> (noting also that <span class="math-container">$\pi_{t+1}$</span> is <span class="math-container">$\mathcal{F}_{t + 1}$</span>-measurable) to get the expression you are looking for.</p>
<p>Hope this helps!</p>

CC BY-SA 4.0
2023-06-10T14:46:47.363Z
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
CC BY-SA 4.0
2023-06-10T14:55:08.713Z
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
CC BY-SA 4.0
2023-06-11T09:19:36.737Z
@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.
Wittgenstein's Poker
CC BY-SA 4.0
2023-06-11T19:45:30.420Z
Woops yeah my mistake but point still stands and thank you.
Econ
CC BY-SA 4.0
2023-06-07T10:34:05.720Z
2023-06-09T10:46:50.133Z
2023-06-09T10:46:50.133Z
Wittgenstein's Poker
<p>In Cochrane's &quot;Fiscal Theory of the Price Level&quot;, I am struggling with the following derivation. Take the first line as given, where <span class="math-container">$\pi_t$</span> and <span class="math-container">$i_t$</span> are random processes adapted to <span class="math-container">$\mathcal{F}_t$</span>; the second line supposedly follows. <span class="math-container">$L$</span> is the lag operator, so <span class="math-container">$L x_t = x_{t-1}$</span> and <span class="math-container">$L^{-1}x_t = x_{t+1}$</span> for any random process <span class="math-container">$x_t$</span>.</p>
<p>The problematic implication is the following:</p>
<p><span class="math-container">$$E_t \Big[ (1- \lambda_1^{-1} L)(1 - \lambda_2 L^{-1}) \pi_{t+1} \Big] = \sigma \kappa \lambda_1^{-1} i_t \\ $$</span></p>
<p><span class="math-container">$$ \Rightarrow \pi_{t+1} = E_{t+1} \frac{\lambda_1^{-1}}{(1- \lambda_1^{-1} L)(1 - \lambda_2 L^{-1})} \sigma \kappa i_t + \frac{1}{(1-\lambda_1^{-1}L)} \delta_{t+1},$$</span></p>
<p>where <span class="math-container">$\delta_{t+1}$</span> is a random process adapted to <span class="math-container">$\mathcal{F}_{t+1}$</span> with <span class="math-container">$E_t[\delta_{t+1}] = 0$</span> for all <span class="math-container">$t$</span>.</p>
<p>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 &quot;through the <span class="math-container">$E_t$</span>&quot; because of the law of iterated expectation <span class="math-container">$E_t[E_{t+1}] = E_t$</span>. I believe I can show this, and given this result I can derive the intermediate step
<span class="math-container">$$ E_t \Big[ (1 - \lambda_1^{-1} L) \pi_{t+1} \Big] = E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1}  \sigma \kappa i_t \Big].$$</span> I can attempt to derive the rest directly, but I arrive at a different result. Specifically, I get</p>
<p><span class="math-container">$$ E_t \pi_{t+1} = E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1}  \sigma \kappa i_t \Big] + \lambda_1^{-1} \pi_t $$</span></p>
<p><span class="math-container">$$ \Rightarrow \pi_{t+1} = E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1}  \sigma \kappa i_t \Big] + \lambda_1^{-1} \pi_t + \delta_{t+1},$$</span></p>
<p>where <span class="math-container">$\delta_{t+1}$</span> is just the expectation error <span class="math-container">$\pi_{t+1} - E_t\pi_{t+1}$</span> and so has the required properties. Iteratively substituting for <span class="math-container">$\pi_t$</span> backwards using this equation yields</p>
<p><span class="math-container">$$ \pi_{t+1} = (1 - \lambda_1^{-1} L)^{-1} E_t \Big[ \lambda_1^{-1} (1 - \lambda_2 L^{-1})^{-1}  \sigma \kappa i_t \Big] + (1 - \lambda_1^{-1} L)^{-1} \delta_{t+1}.$$</span></p>
<p>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 <span class="math-container">$E_{t+1}$</span> comes from.</p>
<p><strong>UPDATE:</strong></p>
<p>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</p>
<p><span class="math-container">$$
-\gamma = (1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1} - E_{t+1} \Big[(1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1} \Big].
$$</span></p>
<p>This means that <span class="math-container">$\gamma$</span> is defined as negative one times the expectation error of <span class="math-container">$(1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1}) \pi_{t+1}$</span> relative to its expectation at time <span class="math-container">$t+1$</span>. We can expand the lag polynomials,
<span class="math-container">\begin{align}
(1-\lambda_1^{-1}L)(1 - \lambda_2 L^{-1})\pi_{t+1} &amp;= \Big( 1 + \lambda_1^{-1} \lambda _2 - \lambda_1^{-1} L - \lambda_2 L^{-1} \Big) \pi_{t+1}  \\
&amp;= (1 + \lambda_1^{-1} \lambda_2) \pi_{t+1} - \lambda_1^{-1} \pi_t - \lambda_2 \pi_{t+2} \\
&amp;\equiv \xi_{t+2},
\end{align}</span>
where I've put a <span class="math-container">$t+2$</span> subscript on <span class="math-container">$\xi$</span> since it is in the <span class="math-container">$\mathcal{F}_{t+2}$</span> information set and not in <span class="math-container">$\mathcal{F}_{t+1}$</span>, as it is a function only of random variables which themselves are in this information set and at least one of these variables (<span class="math-container">$\pi_{t+2}$</span>) is not in <span class="math-container">$\mathcal{F}_{t+1}$</span>. Substituting into the equation for <span class="math-container">$\gamma$</span>, we get</p>
<p><span class="math-container">\begin{align}
-\gamma &amp;= \xi_{t+2} - E_{t+1} \xi_{t+2}.
\end{align}</span>
For the same reason as for <span class="math-container">$\xi_{t+2}$</span>, it seems to me that <span class="math-container">$\gamma$</span> must also be in the <span class="math-container">$\mathcal{F}_{t+2}$</span> information set and not in the <span class="math-container">$\mathcal{F}_{t+1}$</span> information set.</p>
<p>Therefore,
<span class="math-container">\begin{align}
E_{t+1} \frac{\gamma_{t+2}}{(1-\lambda_1^{-1}L)(1-\lambda_2 L^{-1})} \ne 0
\end{align}</span>
in general. The terms involving <span class="math-container">$E_{t+1} \gamma_{t+2}$</span> are zero, as are the terms involving <span class="math-container">$E_{t+1} \gamma_{t+2+j} = E_{t+1} \big[ E_{t+1+j} \gamma_{t+2+j} \big] = 0$</span> for <span class="math-container">$j &gt; 0$</span>. But there are also terms with <span class="math-container">$\gamma_{t+2-j}$</span> for <span class="math-container">$j &gt; 0$</span>, which are in the <span class="math-container">$\mathcal{F}_{t+2-j}$</span> information set and hence are known at time <span class="math-container">$t+1$</span> and therefore their expectation at time <span class="math-container">$t+1$</span> is not equal to zero (i.e., not equal to the function that maps all <span class="math-container">$\omega \in \Omega$</span> to zero, where <span class="math-container">$\Omega$</span> is the underlying state space).</p>
<p>I don't think this would change the overall point that inflation <span class="math-container">$\pi_t$</span> is a weighted average of forward and past nominal interest rates <span class="math-container">$i_t$</span> plus some weighted average of past expectation errors. It would just add another weighted average of a different set of past expectation errors.</p>
<p>Finally, another angle on this is
<span class="math-container">\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}</span>
for <span class="math-container">$t' \ne t$</span>.</p>

CC BY-SA 4.0
2023-06-09T13:09:06.267Z
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?).
Richard Hardy
CC BY-SA 4.0
2023-06-10T14:42:57.497Z
Yes, that's the one. Idk, I think of it as a textbook, but I guess that's up to interpretation.
Econ
<macroeconomics><econometrics><mathematical-economics><applied-econometrics><difference-equations>
Solving stochastic difference equation in New Keynesian model (FTPL textbook derivation)
Econ
# Solving stochastic difference equation in New Keynesian model (FTPL textbook derivation)

(Asked by: username_0 on 2023-06-04)

In username_2'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 username_2 username_2'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 $t' \ne t$.

**Tags:** macroeconomics, econometrics, mathematical-economics, applied-econometrics, difference-equations

> **Comments:**
> **username_1**: What is FTPL textbook? My first guess would be username_2's "Fiscal Theory of the Price Level", but that one is not a textbook (or is it?).  
> **username_0**: Yes, that's the one. Idk, I think of it as a textbook, but I guess that's up to interpretation.  

---

## 2 answers

### Answer by username_2 on 2023-06-15. Score: 3

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}

username_2 username_2


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### Answer by username_3 on 2023-06-07. Score: 1

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!

> **Comments:**
> **username_0**: 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!!  
> **username_0**: Sorry should have written "is not adapted to $\mathcal{F}_{t+1}$", since it involves an infinite series of forward lags of $\pi_{t+1}$.  
> **username_3**: @username_0 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.  
> **username_0**: Woops yeah my mistake but point still stands and thank you.  

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