Solutions: Conditional vs unconditional expectations in RE models
Macroeconomics (M8674), April 2026
Vivaldo Mendes, ISCTE
vivaldo.mendes@iscte-iul.pt
An uncoupled model with Rational Expectations
Consider a simple model described by the following three equations:
\[y_t =\beta \mathbb{E}_t y_{t+1}+x_t \qquad \qquad \qquad \qquad \qquad \qquad \tag{1}\] \[x_t =\phi+\rho x_{t-1}+\varepsilon_t \ \ , \quad \varepsilon_t \sim \mathcal{N}\left(0, \sigma^2\right) \qquad \tag{2}\]
\[z_t =\sigma + \mu y_t \qquad \qquad \qquad \qquad \qquad \qquad \qquad \tag{3}\] where \(\{\beta,\phi,\rho ,\sigma,\mu \}\) are parameters.
- What kind of variables (forward-looking, backward-looking, and static) do we have in this model?
- To secure one stable solution for this model, what are the constraints that we have to impose upon the parameters?
- Solve for the model’s deterministic steady-state (or long-term equilibrium).
Given the following parameters, what are the long-term equilibrium levels of \(y_{t}\), \(x_{t}\) and \(z_t\), according to the hypothesis of conditional expectations? \[\beta=0.75 \ , \ \phi=10 \ , \ \rho=0.5 \ , \ \sigma = 2 \ , \ \mu=0.1 \]
Now consider that the system is in its long-term equilibrium. If in a given period \(t\), \(x_t\) suffers a shock equal to \(\varepsilon_t=+1\) (no more shocks afterward), what happens to \(x_t,\) \(y_t,\) and \(z_t\)? And what will their values be in \(t+1\)?
Considering the same shock and the same parameters as above, what happens to \(y_{t}\), \(x_{t}\), and \(z_t\), according to the hypothesis of unconditional expectations?
When will the two solutions (under conditional and unconditional expectations) be the same?
Solutions
- What kind of variables (forward-looking, backward-looking, and static) do we have in this model?
Solution 1.
This is a dynamic model and has three variables: \(y_t\) (a forward-looking variable); \(x_t\) (a backward-looking variable also called as predetermined), and \(z_t\) which is a static variable (no dynamics in itself).
- To secure one stable solution, what are the constraints that we have to impose upon the parameters?
Solution 2.
As we have a dynamic model, its stability depends on the eigenvalues associated with the system. As this model has uncoupled blocks (each equation can be solved separately), the eigenvalues correspond to individual parameters: \(\beta\) in eq. (1) and \(\rho\) in eq. (2). Eq. (3) represents a static variable (\(z_t\)), and static variables have no internal dynamics, and so the parameters in eq. (3) are irrelevant to the stability of this model.
Therefore, to secure a unique and stable equilibrium in this model, we have to impose the following conditions:
\[|\beta|< 1 \ , \ |\rho| < 1\]
- Solve for the model’s deterministic steady-state (or long-term equilibrium).
Solution 3.
The deterministic solution of the model is the solution without the random component (\(\varepsilon_t=0, \forall t\)). It is as if no random factors were affecting the model’s dynamics. Therefore, to solve for the deterministic steady state, we have to impose the following conditions:
\[x_t=x_{t-1}=\overline{x} \ , \ y_t = \mathbb{E}_t y_{t+1} = \overline{y} \ , \ z_t=z_{t-1}=\overline{z}\]
Applying those conditions to each of the three equations above, we get:
Solution 3 (continuation)
\[\bar{x}=\phi+\rho \bar{x}+0 \ \Rightarrow \ \bar{x}(1-\rho)=\phi \ \Rightarrow \ \bar{x}=\frac{\phi}{1-\rho}\tag{4}\]
\[\bar{y}=\beta \bar{y}+\bar{x} \ \Rightarrow \ \bar{y}(1-\beta)=\bar{x} \ \Rightarrow \ \bar{y}=\frac{\bar{x}}{1-\beta}=\frac{\phi /(1-\rho)}{1-\beta} \tag{5}\]
\[\bar{z}=\sigma+\mu \bar{y} \ \Rightarrow \ \bar{z}=\sigma+\frac{\mu \phi /(1-\rho)}{1-\beta} \tag{6}\]
- Given the following parameters, what are the long-term equilibrium levels of \(y_{t}\), \(x_{t}\) and \(z_t\), according to the hypothesis of conditional expectations? \[\beta=0.75 \ , \ \phi=10 \ , \ \rho=0.5 \ , \ \sigma = 2 \ , \ \mu=0.1\]
Solution 4.
Let us start with eq. (2). Iterate backward in time up to the 3rd iteration, then generalize to the \(n\)-th iteration. Assuming that \(|\rho| < 1\) in order to rule out explosive behavior, we get (details in Appendix A):
Solution 4 (continuation).
\[ x_t=\phi+\rho x_{t-1}+\varepsilon_t \ \ \Rightarrow \ \ x_t= \underbrace{\frac{\phi}{1-\rho}}_{deterministic} + \underbrace{\sum_{i=0}^{n-1} \rho^i \varepsilon_{t-i}}_{random \ part} \tag{7} \]
If \(\phi=10 \ , \rho=0.5\), then
\[ x_t=\underbrace{\frac{10}{1-0.5}}_{determinist\ part} + \ \ \underbrace{\sum_{i=0}^{n-1} 0.5^i \varepsilon_{t-i}}_{random \ part} \ = \ \underbrace{20}_{\bar{x}} \ + \ \underbrace{\sum_{i=0}^{n-1} 0.5^i \varepsilon_{t-i}}_{x^\varepsilon} \tag{8} \] Therefore, we can simply write the solution to \(x_t\) as the sum of a deterministic part \((\bar{x})\) and a a random component \((x^\varepsilon)\): \[x_t=\bar{x}+x^\varepsilon \tag{8a}\]
Solution 4 (continuation).
It is easy to see that the solution in (8)-(8a) depends on the shocks that may hit this process over time. - If we have no shocks, \(x^\varepsilon=0\), and we are back to the deterministic case where \(x_t = \bar{x}=20\) - If we have one shock, we will see what happens in question 5 - If we have many shocks, it is better to use the computer
Now, let us move on to eq. (1). Iterate (now) forward up to the 3rd iteration, then generalize to the \(n\)-th iteration. Assuming that \(\beta< |1|\) in order to rule out explosive behavior, we get (details in Appendix B):
\[ y_t=\beta \mathbb{E}_t y_{t+1}+x_t \ \ \Rightarrow \ \ y_t=\sum_{i=0}^{n-1} \beta^i \mathbb{E}_t x_{t+i} \tag9 \] But, in eq. (9), what is the value of \(\mathbb{E}_t x_{t+i}\)? Considering that in eq. (2) we have \(x_t=\phi+\rho x_{t-1}+\varepsilon_t\), then we will get (details in Appendix C)
\[ \mathbb{E}_t x_{t+i}= \underbrace{\frac{\phi}{1-\rho}}_{deterministic}+ \underbrace{\rho^i x_t^\varepsilon}_{random} \tag{10} \]
Inserting eq. (10) into (9), we get:
\[ \begin{aligned} y_t=\sum_{i=0}^{n-1} \beta^i \mathbb{E}_t x_{t+i} & =\sum_{i=0}^{n-1} \beta^i\left(\frac{\phi}{1-\rho}+\rho^i x^\varepsilon_t\right)= \underbrace{\sum_{i=0}^{n-1} \beta^i\left(\frac{\phi}{1-\rho}\right)}_{deterministic}+ \underbrace{\sum_{i=0}^{n-1} (\beta \rho)^i x^\varepsilon_t}_{random} \end{aligned} \]
Now, we must obtain the solution to the two geometric sums above. details in Appendix D to get the details associated with a geometric sum and its solution. The result will come out as follows:
\[ y_t= \underbrace{\frac{\phi /(1-\rho)}{1-\beta}}_{deterministic}+\underbrace{\frac{x^\varepsilon_t}{1-\beta \rho}}_{random} \tag{11} \]
Taking into account that \(\phi=10, \rho=0.5, \beta = 0.75\), the solution to eq. (11) is given by:
\[ y_t=\frac{10 /(1-0.5)}{1-0.75}+\frac{x^\varepsilon_t}{1-0.75\times 0.5} = \underbrace{80}_{determ.} + \underbrace{1.6 x^\varepsilon_t}_{random} \tag{12} \]
It is easy to see what forces affect the determination of \(y_t\):
- The first one is the impact of the deterministic value of \(x_t\), which is \((\overline{x}= 10/(1-0.5)=20)\). This impact is given by the following term in eq. (12): \[\frac{10 /(1-0.5)}{1-0.75}=80\]
- The second one is the (indirect) impact that the shock exerts upon \(y_t\), given by \(1.6x^\varepsilon_t\). Graphically, this impact can be represented as: \(\varepsilon_t \rightarrow x^\varepsilon_t \underbrace{ \rightarrow y_t}_{1.6x^\varepsilon_t}\)
- If there are no shocks, we will have \(x^\varepsilon_t=0\), and \(y_t=\overline{y}=80\). We are back to the deterministic solution.
- Now consider that the system is in its long-term equilibrium. If in a given period \(t\), \(x_t\) suffers a shock equal to \(\varepsilon_t=+1\) (no more shocks afterward), what happens to \(x_t,\) \(y_t,\) and \(z_t\)? And what will their values be at \(t+1\)?
Solution 5
Here, we have to start with \(x_t\) because this variable directly suffers the shock. The dynamics of \(x_t\) is given by eq. (8) above. Let us bring it back:
\[ x_t=\underbrace{20}_{\text {deterministic }}+\underbrace{\sum_{i=0}^{n-1} 0.5^i \varepsilon_{t-i}}_{\text {random part }} \tag{8a} \]
And the shock is this:
\[\varepsilon_t=+1 \ , \ \varepsilon_{t+i}=0\]
If the shock occurs at period \(t\), let us iterate the model forward two times: \(i=0,1,2\), and see what happens to our variable \(x\) over time \((x_{t+i})\). Notice that \(t+0\) is when the shock occurs (the initial period).
\[
\begin{array}{ll}
i=0 \Rightarrow x_{t+0}=20+0.5^i \varepsilon_{t+0-i} =20+0.5^0 \times \varepsilon_{t+0-0} = \ \ \quad20+\qquad 0.5^0 \times 1=21 \\
i=1 \Rightarrow x_{t+1}=20+0.5^i \varepsilon_{t+1-i}=20+0.5^1 \times \varepsilon_{t+1-1}= \ \ \quad 20
+ \qquad 0.5^1 \times 1=20.5 \\
i=2 \Rightarrow x_{t+2}=20+0.5^i \varepsilon_{t+2-i}=20+0.5^2 \times \varepsilon_{t+2-2}=
\underbrace{20}_{deterministic} + \ \underbrace{0.5^2 \times 1}_{random}=20.25
\end{array}
\]
As expected, the deterministic part of \(x_t\) remains constant \((\overline{x}=20)\). The change occurs in the random part of this process \((x^\varepsilon_{t+i})\):
\[x_t = 20+\underbrace{1}_{x^\varepsilon_{t}} \ , \ x_{t+1} = 20+\underbrace{0.5}_{x^\varepsilon_{t+1}} \ \ , \ x_{t+2} = 20+\underbrace{0.25}_{x^\varepsilon_{t+2}} \ \ , ...\]
As we know the relationship between \(y_t\) and \(x_t\) that solves the model, we can obtain the expected impact of such shock upon \(y_t\). That relationship was given above by eq. (12). Let us bring it back:
\[ y_t=\underbrace{80}_{\text {determ. }}+\underbrace{1.6 x^\varepsilon_t}_{\text {random }} \tag{12a} \]
As the deterministic part of \(x\) does not change, only its random part does, it is immediate to get that:
\[
\begin{array}{ll}
y_{t+0}=80+1.6\times1 \ \ \ \ = \ 81.6 \ \\
y_{t+1}=80+1.6\times 0.5 \ = \ 80.8 \ \ \\
y_{t+2}=80+1.6\times 0.25=80.4
\end{array}
\]
The shock that hits \(x_t\) has a persistent impact over time, indirectly affecting \(y_t\) as well.
Now, consider the static variable (eq. 3):
\[z_t = 2+0.1y_t\]
It is immediate to see the impact of the shock \(\varepsilon_t=+1\) upon \(z_t\):
\[ \begin{array}{ll} z_{t+0}=2+0.1 \times 81.6 = 10.16 \\ z_{t+1}=2+0.1\times 80.8 = 10.08 \\ z_{t+2}=2+0.1\times 80.4=10.04 \end{array} \]
- Considering the same shock and the same parameters as above, what happens to \(y_{t}\), \(x_{t}\), and \(z_t\), according to the hypothesis of unconditional expectations?
Solution 6
- With unconditional expectations, everything becomes easier to calculate.
- The solution to \(x_t\) will be the same (eq. 8):
\[ x_t=\underbrace{20}_{\text {deterministic}}+\underbrace{\sum_{i=0}^{n-1} 0.5^i \varepsilon_{t-i}}_{\text {random part }} \tag{8b} \]
- The solution to \(y_t\) is different in the case of unconditional expectations because, in this case, we care about the unconditional mean of \(x\) in the solution to \(y_t\). This solution was presented above (eq. 9). Let us bring it back:
\[ y_t=\sum_{i=0}^{n-1} \beta^i \mathbb{E}_t x_{t+i} \tag{9b} \]
- What is the unconditional mean of \(x_t\)? It is \(\mathbb{E}_t x_{t+i}= \overline{x} = 20\). details in Appendix B. So, just insert this result into eq. (9b), and we will get:
\[ y_t=\sum_{i=0}^{n-1} \beta^i \mathbb{E}_t x_{t+i}=\sum_{i=0}^{n-1} \beta^i \ \overline{x}=\frac{\overline{x}}{1-\beta} =\frac{20}{1-0.75}=80 \tag{13} \]
The solution for \(z_t\) is immediate: insert the result in eq. (13) into eq. (3):
\[z_t=\sigma + \mu y_t= \sigma + \frac{\mu \overline{x}}{1-\beta}=2+\frac{0.1 \times 20}{1-0.75}=10\]
Solution 6 (continuation)
- Notice that the results in the two previous slides can be summarized as: \[x_t=\overline{x}+\sum_{i=0}^{n-1} 0.5^i \varepsilon_{t-i}=20+\sum_{i=0}^{n-1} 0.5^i \varepsilon_{t-i}\]
\[\ y_t=80 \ , \ z_t = 10 \ , \forall{t}\]
The solutions to \(y_t\) and \(z_t\) are the same as those in the case of the deterministic solution in Question 3: see eqs. (5) and (6).
But they are different from those obtained under conditional expectations, because in the current case we do not take into account the impact from the shocks that affect \(x_t\).
Solution 6 (continuation)
The solution to \(x_t\) is the same under conditional and unconditional expectations because \(x_t\) is a predetermined (or backward-looking) variable.
So, if shocks hit the economy, unconditional expectations will provide a wrong answer to what truly happens in the economy. That is why all modern macroeconomic models use conditional expectations.
- When will the two solutions (under conditional and unconditional expectations) be the same?
Solution 7
Only in the following case: no shocks will hit the economy (model) over time.
Appendices
Each one includes important techniques that are required for solving the questions above.
Appendix A
Solving a backward-looking equation:
\[ \begin{aligned} (iteration \ 1) \rightarrow \ \ x_t & =\phi+\rho x_{t-1}+\varepsilon_t \\ (iteration \ 2) \rightarrow \ \ x_t & =\phi+\rho\left(\phi+\rho x_{t-2}+\varepsilon_{t-1}\right)+\varepsilon_t \\ & =\phi+\rho \phi+\rho^2 x_{t-2}+\rho \varepsilon_{t-1}+\varepsilon_t \\ (iteration \ 3) \rightarrow \ \ x_t & =\phi+\rho \phi+\rho^2\left(\phi+\rho x_{t-3}+\varepsilon_{t-2}\right)+\rho \varepsilon_{t-1}+\varepsilon_t \\ & =\phi+\rho \phi+\rho^2 \phi+\rho^3 x_{t-3}+\rho^2 \varepsilon_{t-2}+\rho \varepsilon_{t-1}+\varepsilon_t \\ & =\sum_{i=0}^{3-1} \rho^i \phi+\rho^3 x_{t-3}+\sum_{i=0}^{3-1} \rho^i \varepsilon_{t-i} \\ (iteration \ n) \rightarrow \ \ x_t & =\sum_{i=0}^{n-1} \rho^i \phi+ \color{blue}{}\rho^n x_{t-n}\color{black}{}+\sum_{i=0}^{n-1} \rho^i \varepsilon_{t-i}\\ & =\sum_{i=0}^{n-1} \rho^i \phi+\sum_{i=0}^{n-1} \rho^i \varepsilon_{t-i} \ \ , \color{blue}{} \ \ \ \ \text{as} \ \ \ \rho< |1| \Rightarrow \rho^n x_{t-n} \rightarrow 0. \qquad \qquad \end{aligned} \]
Appendix B
Solving a forward-looking equation:
\[ \begin{aligned} (iteration \ 1) \rightarrow \ \ y_t & =\beta \mathbb{E}_t y_{t+1}+x_t \\ (iteration \ 2) \rightarrow \ \ y_t & =\beta\left(\beta \mathbb{E}_t y_{t+2}+\mathbb{E}_t x_{t+1}\right)+x_t \\ & =\beta^2 \mathbb{E}_t y_{t+2}+\beta \mathbb{E}_t x_{t+1}+x_t \\ (iteration \ 3) \rightarrow \ \ y_t & =\beta^2\left(\beta \mathbb{E}_t y_{t+3}+\mathbb{E}_t x_{t+2}\right)+\beta \mathbb{E}_t x_{t+1}+x_t \\ & =\beta^3 \mathbb{E}_t y_{t+3}+\beta^2 \mathbb{E}_t x_{t+2}+\beta \mathbb{E}_t x_{t+1}+x_t \\ & =\beta^3 \mathbb{E}_t y_{t+3}+\sum_{i=0}^{3-1} \beta^i \mathbb{E}_t x_{t+i} \\ (iteration \ n) \rightarrow \ \ y_t & = \color{blue}{}\beta^n \mathbb{E}_t y_{t+n} \color{black}{}+\sum_{i=0}^{n-1} \beta^i \mathbb{E}_t x_{t+i} \\ & =\sum_{i=0}^{n-1} \beta^i \mathbb{E}_t x_{t+i} \ \ , \quad \color{blue}{} \text{as} \ \ |\beta|< 1 \Rightarrow \beta^n \mathbb{E}_t y_{t+n} \rightarrow 0. \qquad \qquad \end{aligned} \]
Appendix C
Conditional expectations:
\[x_{t}=\phi + \rho x_{t-1}+\varepsilon_{t} \ \ , \quad \varepsilon_t \sim \mathcal{N}\left(0, \sigma^2\right) \tag{A1}\]
To compute the conditional expectation of \(x_{t+i}\), \((\mathbb{E}_t x_{t+i})\), apply the expectations operator to eq. (A1), up to the third iteration:
$$ \[\begin{aligned} & \qquad x_t= \phi+\rho x_{t-1}+\varepsilon_t \\ & \mathbb{E}_t x_{t+1}=\phi+\rho \mathbb{E}_t x_t+\mathbb{E}_t \varepsilon_{t+1}\ \ \ =\phi+\rho x_t+0 \qquad \qquad \qquad \ =\phi+\rho x_t \\ & \mathbb{E}_t x_{t+2}=\phi+\rho \mathbb{E}_t x_{t+1}+\mathbb{E}_t \varepsilon_{t+2}=\phi+\rho\left[\phi+\rho x_t\right]+0 \qquad \quad =\phi+\rho \phi+\rho^2 x_t\\ &\mathbb{E}_t x_{t+3}=\phi+\rho \mathbb{E}_t x_{t+2}+\mathbb{E}_t \varepsilon_{t+3}=\phi+\rho\left[\phi+\rho \phi+\rho^2 x_t\right] +0 = \underbrace{\phi+\rho \phi+\rho^2 \phi}_{=\sum_{k=0}^{3-1} \phi \rho^k}+ \rho^3x_t \end{aligned}\]\[ That is, at the 3rd iteration, we get: \]t x{t+3}=_{k=0}^{3-1} ^k +^3 x_t$$
Generalizing to the \(i\)th iteration, we obtain:
\[\mathbb{E}_t x_{t+i}=\sum_{k=0}^{i-1} \phi \rho^k +\rho^i x_t=\frac{\phi}{1-\rho}+\rho^{i} x_t \tag{A2}\]
But as the deterministic mean of \(x_t\) is given by \(\bar{x}=\frac{\phi}{1-\rho}\), then we can rewrite (A2) as:
\[\mathbb{E}_t x_{t+i}=\underbrace{\frac{\phi}{1-\rho}}_{\bar{x}}+\underbrace{\rho^i x_t}_{x^\varepsilon}=\bar{x}+ \rho^i x_t^\varepsilon \tag{A3}\]
where \(x^\varepsilon\) is the random component of this process affecting \(x_t\).
Appendix D
Solution of a Geometric Series:
- Suppose we have a process that is written as: \[ s=\rho^0 \phi+\rho^1 \phi +\rho^2 \phi +\rho^3 \phi \ + \ ... \ = \sum_{i=0}^{\infty} \phi \rho^i \]
- It has two crucial elements:
- First term of the series (when \(i=0\)): \(\ \phi\)
- The common ratio: \(\rho\)
- The solution is given by the expression: \[s=\frac{\text {first term }}{1-\text { common ratio }}=\frac{\phi}{1-\rho} \ \ , \qquad if \ \ |\rho|<1\]
- No solution: if \(|\rho| > 1\) , \(s\) is explosive; if \(|\rho| = 1\), \(s\) does not converge to any value.
Appendix E
Unconditional expectations:
\[x_{t}=\phi + \rho x_{t-1}+\varepsilon_{t} \ \ , \quad \varepsilon_t \sim \mathcal{N}\left(0, \sigma^2\right)\]
- Assuming unconditional expectations, the mean is given by the (deterministic) steady-state value of \(x_t\):
\[x_{t}= x_{t-1}= \overline{x}\]
Which leads to: \[\overline{x}=\phi + \rho \overline{x}+0 \Rightarrow \overline{x}=\frac{\phi}{ 1-\rho} \quad , \qquad \rho \neq 1. \]
Therefore, the expected (unconditional) value of \(\mathbb{E}_t x_{t+i}\) is given by: \[\mathbb{E}_t x_{t+i}=\overline{x}=\frac{\phi}{ 1-\rho} \tag{A4}\]