<p>Presumably <span class="math-container">$I_t$</span> stands for the information set at time <span class="math-container">$t$</span>. So note that <span class="math-container">$E[\epsilon_j|I_t] = \epsilon_j$</span> <span class="math-container">$\forall j \leq t$</span>. This will come into play when working out the LHS.</p>

<p>We wish to verify Equation (4), which is </p>

<p><span class="math-container">$$
E(p_{t+1}|I_t) = \frac{\Psi(B) - \psi_0}{B} \epsilon_t,
$$</span></p>

<p>which is equivalent to</p>

<p><span class="math-container">$$
$$</span></p>

<p><span class="math-container">$$
\begin{align*}
B(E(p_{t+1}|I_t)) = \Psi(B) \epsilon_t - \psi_0 \epsilon_t. &amp;&amp;(*)
\end{align*}
$$</span></p>

<p><span class="math-container">$$
$$</span></p>

<hr>

<p>We are given that</p>

<p><span class="math-container">$$
\Psi(B) \epsilon_t = \psi_0 \epsilon_t + \psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_k \epsilon_{t-k} + \dots
$$</span></p>

<p>so the <strong>RHS</strong> of <span class="math-container">$(*)$</span> is</p>

<p><span class="math-container">$$
$$</span></p>

<p><span class="math-container">$$
\boldsymbol{\Psi(B) \epsilon_t - \psi_0 \epsilon_t = \psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_{k} \epsilon_{t-k} + \dots}.
$$</span></p>

<p><span class="math-container">$$
$$</span></p>

<hr>

<p>Now, the LHS. Since <span class="math-container">$p_t = \psi_0 \epsilon_t + \psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_k \epsilon_{t-k} + \dots,$</span></p>

<p><span class="math-container">$$
$$</span></p>

<p><span class="math-container">$$
\begin{align*}
E(p_{t+1}|I_t) &amp;= \psi_0 E(\epsilon_{t+1}|I_t) + \psi_1 E(\epsilon_t|I_t) + \psi_2 E(\epsilon_{t-1}|I_t) + \dots  + \psi_k E(\epsilon_{t-k+1}|I_t) + \dots \\
&amp;= \psi_1 \epsilon_{t} + \psi_2 \epsilon_{t-1} + \dots + \psi_{k} \epsilon_{t-k+1} + \dots
\end{align*}
$$</span></p>

<p><span class="math-container">$$
$$</span></p>

<p>since <span class="math-container">$E(\epsilon_{t+1}|I_t) = 0$</span> because <span class="math-container">$\epsilon$</span> is an independent white noise process.</p>

<p>Given that <span class="math-container">$B(\cdot)$</span> is the <em>lag operator</em>, the <strong>LHS</strong> of <span class="math-container">$(*)$</span> is thus</p>

<p><span class="math-container">$$
$$</span></p>

<p><span class="math-container">$$
\begin{align*}
\boldsymbol{B(E(p_{t+1}|I_t))} &amp;= B(\psi_1 \epsilon_{t} + \psi_2 \epsilon_{t-1} + \dots + \psi_{k} \epsilon_{t-k+1} + \dots) \\
&amp;= \psi_1 B(\epsilon_{t}) + \psi_2 B(\epsilon_{t-1}) + \dots + \psi_{k} B(\epsilon_{t-k+1}) + \dots \\
&amp;= \boldsymbol{\psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_{k} \epsilon_{t-k} + \dots}.
\end{align*}
$$</span></p>

<p><span class="math-container">$$
$$</span></p>

<hr>

<p>Thus the <strong>LHS = RHS</strong>; that is,</p>

<p><span class="math-container">$$
\begin{align*}
E(p_{t+1}|I_t) = \frac{\Psi(B) - \psi_0}{B} \epsilon_t. &amp;&amp;\blacksquare
\end{align*}
$$</span></p>

CC BY-SA 4.0
2018-10-18T08:03:23.167Z
Hi Kenneth.  I didn't go over your solution  yet ( too late and brain is closed for the night )  but I'm checking  the answer anyway and  will print out and read carefully. Thanks a lot.
mark leeds
CC BY-SA 4.0
2018-10-18T03:31:18.113Z
2018-10-18T04:14:46.847Z
2018-10-18T04:14:46.847Z
Kenneth Rios
<p>Hi: I'm reading a paper by Broze and Szafarz titled "On Econometric Models with Rational Expectations and I don't follow a step in section 3.2 which discusses the method of undetermined coefficients. So,  I was hoping that someone could explain the derivation. I will write down the equations and then explain the step that I don't follow.</p>

<p>1) <span class="math-container">$p_{t} = a \times E(p_{t+1} | I_{t}) + u_{t} $</span> This equation  is given. It comes out of the Cagan Hyperinflation model. <span class="math-container">$u_{t}$</span> is an error term.</p>

<p>2) <span class="math-container">$ u_{t} = \epsilon_{t} + \alpha \epsilon_{t-1} + \ldots + \alpha_{k} \epsilon_{t-k} + \ldots = A(B) \epsilon_{t} $</span></p>

<p>3) <span class="math-container">$ p_{t} = \psi_{0} \epsilon_{t} + \ldots \psi_{k} \epsilon_{t-k} + \ldots = \Psi(B) \epsilon_{t} $</span></p>

<p>4) <span class="math-container">$ E(p_{t+1}| I_{t} ) = \frac{\Psi(B) - \psi_{0}}{B}    \epsilon_{t} $</span>
So, <span class="math-container">$B$</span> denotes the lag operator which is often denoted as <span class="math-container">$L$</span> in the literature. The explanations of the equations follow.</p>

<p>1) As stated above, this is given. This is clear.</p>

<p>2) This is the standard infinite MA representation that follows from Wold's decomposition. This is clear.</p>

<p>3) This is the guess that is made in the method of undetermined coefficients. This is clear.</p>

<p>4) This is not clear to me. They say "it follows". I see that it's kind of coming by subtracting <span class="math-container">$\psi_{0} \epsilon_{t}$</span> from both sides of 3) but <span class="math-container">$B$</span> is the infinite lag polynomial and they are dividing by it which confuses me.</p>

<p>Thanks in advance for the help. Some discussion has been omitted so, if anyone needs to see the actual pages, I can scan and send them. My email is my name with a 2 on the end of it and the mail server is that famous one that starts with a g.</p>

<pre><code>                                        Mark
</code></pre>

CC BY-SA 4.0
2018-10-18T02:39:10.997Z
Equation (4) should be $E(p_{t+1}|I_t) = \frac{\Psi(B) - \psi_0}{B} \epsilon_t$ instead right? In other words, $\Psi$ is a vector $\psi_k$ for $k = 0, \dots, \infty$?
Kenneth Rios
CC BY-SA 4.0
2018-10-18T02:42:47.100Z
thanks. I edited a lot because I figured out part of it. help is still appreciated.
mark leeds
<econometrics><rational-expectations>
not following a step in a paper on rational expectations
mark leeds
# not following a step in a paper on rational expectations

(Asked by: username_0 on 2018-10-18)

Hi: I'm reading a paper by Broze and Szafarz titled "On Econometric Models with Rational Expectations and I don't follow a step in section 3.2 which discusses the method of undetermined coefficients. So,  I was hoping that someone could explain the derivation. I will write down the equations and then explain the step that I don't follow.
1) $p_{t} = a \times E(p_{t+1} | I_{t}) + u_{t} $ This equation  is given. It comes out of the Cagan Hyperinflation model. $u_{t}$ is an error term.
2) $ u_{t} = \epsilon_{t} + \alpha \epsilon_{t-1} + \ldots + \alpha_{k} \epsilon_{t-k} + \ldots = A(B) \epsilon_{t} $
3) $ p_{t} = \psi_{0} \epsilon_{t} + \ldots \psi_{k} \epsilon_{t-k} + \ldots = \Psi(B) \epsilon_{t} $
4) $ E(p_{t+1}| I_{t} ) = \frac{\Psi(B) - \psi_{0}}{B}    \epsilon_{t} $
So, $B$ denotes the lag operator which is often denoted as $L$ in the literature. The explanations of the equations follow.
1) As stated above, this is given. This is clear.
2) This is the standard infinite MA representation that follows from Wold's decomposition. This is clear.
3) This is the guess that is made in the method of undetermined coefficients. This is clear.
4) This is not clear to me. They say "it follows". I see that it's kind of coming by subtracting $\psi_{0} \epsilon_{t}$ from both sides of 3) but $B$ is the infinite lag polynomial and they are dividing by it which confuses me.
Thanks in advance for the help. Some discussion has been omitted so, if anyone needs to see the actual pages, I can scan and send them. My email is my name with a 2 on the end of it and the mail server is that famous one that starts with a g.
                                        Mark

**Tags:** econometrics, rational-expectations
> **Comments:**
> **username_1**: Equation (4) should be $E(p_{t+1}|I_t) = \frac{\Psi(B) - \psi_0}{B} \epsilon_t$ instead right? In other words, $\Psi$ is a vector $\psi_k$ for $k = 0, \dots, \infty$?  
> **username_0**: thanks. I edited a lot because I figured out part of it. help is still appreciated.  

---

## 1 answers

### Answer by username_1 on 2018-10-18. Score: 0 | Accepted answer

Presumably $I\_t$ stands for the information set at time $t$. So note that $E[\epsilon\_j|I\_t] = \epsilon\_j$ $\forall j \leq t$. This will come into play when working out the LHS.

We wish to verify Equation (4), which is

$$
E(p\_{t+1}|I\_t) = \frac{\Psi(B) - \psi\_0}{B} \epsilon\_t,
$$

which is equivalent to

$$
$$

$$
\begin{align\*}
B(E(p\_{t+1}|I\_t)) = \Psi(B) \epsilon\_t - \psi\_0 \epsilon\_t. &&(\*)
\end{align\*}
$$

$$
$$

---

We are given that

$$
\Psi(B) \epsilon\_t = \psi\_0 \epsilon\_t + \psi\_1 \epsilon\_{t-1} + \psi\_2 \epsilon\_{t-2} + \dots + \psi\_k \epsilon\_{t-k} + \dots
$$

so the **RHS** of $(\*)$ is

$$
$$

$$
\boldsymbol{\Psi(B) \epsilon\_t - \psi\_0 \epsilon\_t = \psi\_1 \epsilon\_{t-1} + \psi\_2 \epsilon\_{t-2} + \dots + \psi\_{k} \epsilon\_{t-k} + \dots}.
$$

$$
$$

---

Now, the LHS. Since $p\_t = \psi\_0 \epsilon\_t + \psi\_1 \epsilon\_{t-1} + \psi\_2 \epsilon\_{t-2} + \dots + \psi\_k \epsilon\_{t-k} + \dots,$

$$
$$

$$
\begin{align\*}
E(p\_{t+1}|I\_t) &= \psi\_0 E(\epsilon\_{t+1}|I\_t) + \psi\_1 E(\epsilon\_t|I\_t) + \psi\_2 E(\epsilon\_{t-1}|I\_t) + \dots + \psi\_k E(\epsilon\_{t-k+1}|I\_t) + \dots \\
&= \psi\_1 \epsilon\_{t} + \psi\_2 \epsilon\_{t-1} + \dots + \psi\_{k} \epsilon\_{t-k+1} + \dots
\end{align\*}
$$

$$
$$

since $E(\epsilon\_{t+1}|I\_t) = 0$ because $\epsilon$ is an independent white noise process.

Given that $B(\cdot)$ is the *lag operator*, the **LHS** of $(\*)$ is thus

$$
$$

$$
\begin{align\*}
\boldsymbol{B(E(p\_{t+1}|I\_t))} &= B(\psi\_1 \epsilon\_{t} + \psi\_2 \epsilon\_{t-1} + \dots + \psi\_{k} \epsilon\_{t-k+1} + \dots) \\
&= \psi\_1 B(\epsilon\_{t}) + \psi\_2 B(\epsilon\_{t-1}) + \dots + \psi\_{k} B(\epsilon\_{t-k+1}) + \dots \\
&= \boldsymbol{\psi\_1 \epsilon\_{t-1} + \psi\_2 \epsilon\_{t-2} + \dots + \psi\_{k} \epsilon\_{t-k} + \dots}.
\end{align\*}
$$

$$
$$

---

Thus the **LHS = RHS**; that is,

$$
\begin{align\*}
E(p\_{t+1}|I\_t) = \frac{\Psi(B) - \psi\_0}{B} \epsilon\_t. &&\blacksquare
\end{align\*}
$$

> **Comments:**
> **username_0**: Hi Kenneth.  I didn't go over your solution  yet ( too late and brain is closed for the night )  but I'm checking  the answer anyway and  will print out and read carefully. Thanks a lot.  

---

