<p>Maybe the differencing approach should work for you. Basically the idea is to reduce this series to a finite expression, using the <a href="https://en.wikipedia.org/wiki/Lag_operator" rel="nofollow noreferrer">lag operator</a>. Let me explain with an example from <a href="https://vernonpress.com/book/70" rel="nofollow noreferrer">Costa</a> (2016, p.81) (If you're not familiar with the lag operator I suggest you to read the first section of the page linked above.):</p>
<p>We have the optimal price level equation (equivalent to your NKPC but less developed):</p>
<p><span class="math-container">$$\tilde{P_{t}}=\theta\tilde{P_{t-1}}+(1-\theta)(1-\beta\theta)E_t\sum_{i=0}^{\infty}(\beta\theta)^i\tilde{MC}_{t+i}$$</span></p>
<p>Now, we can see that the difference between one period of time might be represented by:</p>
<p><span class="math-container">$$(\beta\theta)^k\tilde{MC}_{t+k}-(\beta\theta)^{k+1}\tilde{MC}_{t+k+1}=(\beta\theta)^k(\tilde{MC}_{t+k}-\beta\theta\tilde{MC}_{t+k+1})$$</span></p>
<p>In terms of lag operator we could factor <span class="math-container">$\tilde{MC_t}$</span>:</p>
<p><span class="math-container">$$(\beta\theta)^k\tilde{MC}_{t+k}-(\beta\theta)^{k+1}\tilde{MC}_{t+k+1}=(\beta\theta)^k\tilde{MC_t}(1-\beta\theta L^{-1})$$</span></p>
<p>What this tells us is that we can use the same series lagged by <span class="math-container">$-1$</span> (same as forward by 1) to eliminate future periods from the original one. Let <span class="math-container">$S_t=E_t\sum_{i=0}^{\infty}(\beta\theta)^i\tilde{MC}_{t+i}$</span>, then (omitting expectations operator):</p>
<p><span class="math-container">$$S_t-S_{t+1}=
$$</span>
<span class="math-container">\begin{bmatrix}
\tilde{MC}_{t}+(\beta\theta)\tilde{MC}_{t+1}+(\beta\theta)^2\tilde{MC}_{t+2}+(\beta\theta)^3\tilde{MC}_{t+3}+...\\
-\beta\theta\tilde{MC}_{t+1}-(\beta\theta)^2\tilde{MC}_{t+2}-(\beta\theta)^3\tilde{MC}_{t+3}-(\beta\theta)^4\tilde{MC}_{t+4}-...
\end{bmatrix}</span></p>
<p><span class="math-container">$$=\tilde{MC}_t$$</span></p>
<p>Look that we managed to reduce the infinite sum into a single expression, now to replicate this in our prices equation, we need to get something similar to <span class="math-container">$S_t-S_{t+1}$</span> from <span class="math-container">$S_t$</span> only, taking advantage of lags because if we just sum <span class="math-container">$-S_{t+1}$</span> in both sides of the equation we would've done nothing. Then you can check that:</p>
<p><span class="math-container">$$S_t-S_{t+1}=(1-\beta\theta L^{-1})S_t=\tilde{MC}_t$$</span></p>
<p>With that in mind, we just need to multiply both sides of our prices equation by <span class="math-container">$(1-\beta\theta L^{-1})$</span>, and develop the corresponding algebra for arriving to:</p>
<p><span class="math-container">$$\tilde{P_t}-\beta\theta E_t\tilde{P_{t+1}}=(1-\theta)(1-\beta\theta)\tilde{MC}_t-\theta^2\beta\tilde{P}_t+\theta\tilde{P}_{t-1}$$</span></p>
<p>Which allows us to make better analysis by getting rid of the series, and therefore being able to write in terms of matrices and vectors that are more &quot;readable&quot;. I'm pretty sure you can do this same process with your equation that is in term of inflation (instead of prices). Hope this works.</p>

CC BY-SA 4.0
2021-01-24T09:43:34.997Z
Thank you for your answer! This is essentilly equivalent to applying the law of iterated expectations right? My problem , and i should have specified it in the question, is that i am not considering rational expectations. I have not yet specified my expectations operator and for now it is just an empty container. However if i apply the law of iterated expectations i end up with the same IS - NKPC as in the rational expectations model.
qwerty-qwertz
CC BY-SA 4.0
2021-01-24T09:47:40.123Z
i would like to continue to keep the long term expectations in the two equations and not simply the one period ahed forecast (   E_t [ z_{t+1}] ).  However i have to find a way to make the infinite summation more manageble.
qwerty-qwertz
CC BY-SA 4.0
2021-01-24T19:22:09.437Z
Hi: I didn't read it carefully enough to know if it can help you ( or if nrivera already showed this )  but it might. https://francisbach.com/the-sum-of-a-geometric-series-is-all-you-need/
mark leeds
CC BY-SA 4.0
2021-01-23T21:01:25.610Z
2021-01-24T04:04:03.523Z
2021-01-24T04:04:03.523Z
manifold
<p>I have derived the microeconomic foundations of a dsge model and I've obtained the IS and NKPC. I would like to represent them in matrix form to study the system. However the problem is that both equation have in infinite summation. How can i proceed? Do I need to approximate, if possible, the infinite sum to a finite term?</p>
<p><span class="math-container">\begin{equation}
 \tilde \pi_t = E_t \sum^{\infty}_{T=t} (A)^{T-t} \left[ B  x_{T+1} + C \tilde \pi_{T+1} \right]. 
 \end{equation}</span>
<span class="math-container">\begin{equation}
{x}_{t}=\tilde{E}_{t} \sum_{T=t}^{\infty} D^{T-t} 
 \left[F{x}_{T+1}-\sigma\left(\tilde{\imath}_{T}-\tilde{\pi}_{T+1}\right) \right] 
\end{equation}</span></p>

CC BY-SA 4.0
2021-01-23T12:27:17.407Z
Not sure if I understand the question - matrices  can have infinite number of rows or columns- this class of matrices even has its own name - infinite matrices - and most standard matrix operations work/have equivalent with infinite matrices as well (although they can be more difficult). So if the goal is just to represent it as a matrix then there should be no problem you will just have matrix $A_{nm}$ with dimensions going up to infinity e.g. $m=1,2,3... \infty$. $A_{\infty,\infty}$ is valid matrix. Unless you want to run some numeric optimization this does not create problems per se.
1muflon1
CC BY-SA 4.0
2021-01-23T14:18:07.210Z
Hi 1muflon1: I don't think the OP means infinite in the sense you're describing but rather just that the sum is infinite. To the OP: If the terms were scalars less than 1.0, you would use the formula for a geometric series so I have a feeling there's an analogue for matrices ( with some corresponding constraint on the eigenvalues of the matrix ) .  I would check out Magnus and Neudecker or any other text that has all the tricks and closed form expressions for matrices. Or google for "geometric series for matrices"..
mark leeds
CC BY-SA 4.0
2021-02-15T07:35:05.670Z
You can use a value function approach to make the problem recursive or the 'Euler equation method' in which you use the equilibrium conditions to derive a decision rule. This is outlined in [Fernandez-Villaverde et al.](https://doi.org/10.1016/bs.hesmac.2016.03.006)
Joe
<mathematical-economics><dsge><linear-algebra>
How can difference equations with an infinite summation be represented in matrix form?
qwerty-qwertz
# How can difference equations with an infinite summation be represented in matrix form?

(Asked by: username_0 on 2021-01-23)

I have derived the microeconomic foundations of a dsge model and I've obtained the IS and NKPC. I would like to represent them in matrix form to study the system. However the problem is that both equation have in infinite summation. How can i proceed? Do I need to approximate, if possible, the infinite sum to a finite term?

\begin{equation}
\tilde \pi\_t = E\_t \sum^{\infty}\_{T=t} (A)^{T-t} \left[ B x\_{T+1} + C \tilde \pi\_{T+1} \right].
\end{equation}
\begin{equation}
{x}\_{t}=\tilde{E}\_{t} \sum\_{T=t}^{\infty} D^{T-t}
\left[F{x}\_{T+1}-\sigma\left(\tilde{\imath}\_{T}-\tilde{\pi}\_{T+1}\right) \right]
\end{equation}

**Tags:** mathematical-economics, dsge, linear-algebra

> **Comments:**
> **username_1**: Not sure if I understand the question - matrices  can have infinite number of rows or columns- this class of matrices even has its own name - infinite matrices - and most standard matrix operations work/have equivalent with infinite matrices as well (although they can be more difficult). So if the goal is just to represent it as a matrix then there should be no problem you will just have matrix $A_{nm}$ with dimensions going up to infinity e.g. $m=1,2,3... \infty$. $A_{\infty,\infty}$ is valid matrix. Unless you want to run some numeric optimization this does not create problems per se.  
> **username_2**: Hi username_1: I don't think the OP means infinite in the sense you're describing but rather just that the sum is infinite. To the OP: If the terms were scalars less than 1.0, you would use the formula for a geometric series so I have a feeling there's an analogue for matrices ( with some corresponding constraint on the eigenvalues of the matrix ) .  I would check out Magnus and Neudecker or any other text that has all the tricks and closed form expressions for matrices. Or google for "geometric series for matrices"..  
> **username_3**: You can use a value function approach to make the problem recursive or the 'Euler equation method' in which you use the equilibrium conditions to derive a decision rule. This is outlined in [Fernandez-Villaverde et al.](https://doi.org/10.1016/bs.hesmac.2016.03.006)  

---

## 1 answers

### Answer by username_4 on 2021-01-23. Score: 1

Maybe the differencing approach should work for you. Basically the idea is to reduce this series to a finite expression, using the [lag operator](https://en.wikipedia.org/wiki/Lag_operator). Let me explain with an example from [Costa](https://vernonpress.com/book/70) (2016, p.81) (If you're not familiar with the lag operator I suggest you to read the first section of the page linked above.):

We have the optimal price level equation (equivalent to your NKPC but less developed):

$$\tilde{P\_{t}}=\theta\tilde{P\_{t-1}}+(1-\theta)(1-\beta\theta)E\_t\sum\_{i=0}^{\infty}(\beta\theta)^i\tilde{MC}\_{t+i}$$

Now, we can see that the difference between one period of time might be represented by:

$$(\beta\theta)^k\tilde{MC}\_{t+k}-(\beta\theta)^{k+1}\tilde{MC}\_{t+k+1}=(\beta\theta)^k(\tilde{MC}\_{t+k}-\beta\theta\tilde{MC}\_{t+k+1})$$

In terms of lag operator we could factor $\tilde{MC\_t}$:

$$(\beta\theta)^k\tilde{MC}\_{t+k}-(\beta\theta)^{k+1}\tilde{MC}\_{t+k+1}=(\beta\theta)^k\tilde{MC\_t}(1-\beta\theta L^{-1})$$

What this tells us is that we can use the same series lagged by $-1$ (same as forward by 1) to eliminate future periods from the original one. Let $S\_t=E\_t\sum\_{i=0}^{\infty}(\beta\theta)^i\tilde{MC}\_{t+i}$, then (omitting expectations operator):

$$S\_t-S\_{t+1}=
$$
\begin{bmatrix}
\tilde{MC}\_{t}+(\beta\theta)\tilde{MC}\_{t+1}+(\beta\theta)^2\tilde{MC}\_{t+2}+(\beta\theta)^3\tilde{MC}\_{t+3}+...\\
-\beta\theta\tilde{MC}\_{t+1}-(\beta\theta)^2\tilde{MC}\_{t+2}-(\beta\theta)^3\tilde{MC}\_{t+3}-(\beta\theta)^4\tilde{MC}\_{t+4}-...
\end{bmatrix}

$$=\tilde{MC}\_t$$

Look that we managed to reduce the infinite sum into a single expression, now to replicate this in our prices equation, we need to get something similar to $S\_t-S\_{t+1}$ from $S\_t$ only, taking advantage of lags because if we just sum $-S\_{t+1}$ in both sides of the equation we would've done nothing. Then you can check that:

$$S\_t-S\_{t+1}=(1-\beta\theta L^{-1})S\_t=\tilde{MC}\_t$$

With that in mind, we just need to multiply both sides of our prices equation by $(1-\beta\theta L^{-1})$, and develop the corresponding algebra for arriving to:

$$\tilde{P\_t}-\beta\theta E\_t\tilde{P\_{t+1}}=(1-\theta)(1-\beta\theta)\tilde{MC}\_t-\theta^2\beta\tilde{P}\_t+\theta\tilde{P}\_{t-1}$$

Which allows us to make better analysis by getting rid of the series, and therefore being able to write in terms of matrices and vectors that are more "readable". I'm pretty sure you can do this same process with your equation that is in term of inflation (instead of prices). Hope this works.

> **Comments:**
> **username_0**: Thank you for your answer! This is essentilly equivalent to applying the law of iterated expectations right? My problem , and i should have specified it in the question, is that i am not considering rational expectations. I have not yet specified my expectations operator and for now it is just an empty container. However if i apply the law of iterated expectations i end up with the same IS - NKPC as in the rational expectations model.  
> **username_0**: i would like to continue to keep the long term expectations in the two equations and not simply the one period ahed forecast (   E_t [ z_{t+1}] ).  However i have to find a way to make the infinite summation more manageble.  
> **username_2**: Hi: I didn't read it carefully enough to know if it can help you ( or if nrivera already showed this )  but it might. https://francisbach.com/the-sum-of-a-geometric-series-is-all-you-need/  

---

