<p>That's an AR(1) for <span class="math-container">$\lambda(t)$</span> so <span class="math-container">$|y|$</span> has to be less than 1.0 for stationarity.</p>
<p>We have:</p>
<ol>
<li><span class="math-container">$\lambda_{t+1} - y \lambda_{t} = \epsilon_{(t+1)}$</span></li>
</ol>
<p>First, using the lag operator, <span class="math-container">$L$</span>, we can simplify 1. :</p>
<ol start="2">
<li><p><span class="math-container">$\lambda_{t+1}(1 - y L) = \epsilon_{(t+1)}$</span></p>
</li>
<li><p><span class="math-container">$\lambda_{t+1} = \frac{\epsilon_{(t+1)}}{(1 - y L)}$</span></p>
</li>
</ol>
<p>Assuming that <span class="math-container">$0 \lt y \lt  1.0$</span> ( not sure if I need this assumption but, if it's negative, it can't be converted to an infinite geometric series so it seems to me like the assumption is needed ), one can write the denominator as an infinite geometric series: This gives:</p>
<ol start="4">
<li><span class="math-container">$\lambda_{t+1} = \sum_{i=0}^{\infty}{ y^{i} \times \epsilon_{t+1-i}}$</span></li>
</ol>
<p>where <span class="math-container">$0 \lt y  \lt 1.0$</span>.</p>
<p>Now take the expectation of both sides:</p>
<ol start="5">
<li><span class="math-container">$E(\lambda_{t+1}) = \sum_{i=0}^{\infty} y^{i} \times E(\epsilon_{t+1-i})$</span></li>
</ol>
<p>But the expectation of the <span class="math-container">$\epsilon_t$</span> is <span class="math-container">$(1-y)$</span> so 5) becomes:</p>
<ol start="6">
<li><span class="math-container">$E(\lambda_{t+1}) = (1 - y) \sum_{i=0}^{\infty} y^{i} $</span></li>
</ol>
<p>But the geometric series can be re-written as one term which is <span class="math-container">$\frac{1}{1-y}$</span> so we obtain:</p>
<ol start="7">
<li><span class="math-container">$E(\lambda_{t+1}) = \frac{(1-y)}{(1-y)} = 1.0$</span></li>
</ol>
<p>So, the mean of <span class="math-container">$\lambda_{t+1} = 1.0$</span> and  <span class="math-container">$\epsilon_{t}$</span> is always non-negative so 4) implies that <span class="math-container">$\lambda_{t+1}$</span> is always non-negative.</p>

CC BY-SA 4.0
2023-08-02T06:15:12.137Z
Could you please explain Equation (2)? I didn't get how you have factored out λ_t+1 from the last term - what does L denote? And what is the economic intuition of the denominator in Equation (4)?
S.G.
CC BY-SA 4.0
2023-08-02T06:28:17.853Z
$L$ is a lag operator with the definition that $L(x_{t+1}) = x_{t}$ so it lags the time, $t$ by one period.  It's pretty much the same thing as what z does in a z-transform if you are familar with that ?. I'm not a math person so maybe one of the mathematicians on this list can explain it more clearly or in more depth. Hamilton's "Time Series Analysis" also has an explanation.
mark leeds
CC BY-SA 4.0
2023-08-02T06:32:04.783Z
Do you want me to change $y$ to $\rho$ everywhere because using $y$ to me is confusing. It's easy enough to change. Basically, 4) says that $\lambda$ is really just an exponentially smoothed function of the past shocks $\epsilon_t$. The lag on $\lambda_{t+1}$ on the right hand side is gone because the response ( the left hand side )  is only a function of all the past shocks that have happened upto and including time $(t+1)$
mark leeds
CC BY-SA 4.0
2023-08-02T06:36:59.123Z
If you're into macro ( or rational expectations ) or planning on getting into macro or rational expectations, definitely you should familiarize yourself with the lag operator. Macro uses a lot of time series so the lag operator is crucial to understanding the literature. Maybe I should say: "attempting to understand the literature"  because the macro-literature is quite difficult for me. IMHO, Lucas is the clearest writer and maybe the only clear writer ? His "Studies in Business Cycle Theory" has the clearest explanations of RE that I know of but its old so totally RE oriented.
mark leeds
CC BY-SA 4.0
2023-08-02T06:53:34.233Z
One thing that might look confusing is that ( 1 - y L) is in the denominator and I said eartlier that the lag operator lags it input by one. But there's no $t$ on the input $y$. so that seems weird. What's happening there is that the lag operator is actually working on the numerator, $\epsilon_t$. The $y$ is just a constant so nothing happens to that. So, if you had say $\frac{1}{1-p} $ where $p$ was positive and less than 1.0, that expression can be re-written as an infinite series: $1 + p + p^2 + \ldots + p^n$ all the way to $\infty$.
mark leeds
CC BY-SA 4.0
2023-08-02T06:55:06.457Z
The above is analogous to what I'm doing in 4) with the only difference being that the infinite series is $1 + y L +  y^2 L^2 + \ldots + y^n L^{n} $  all the way to $\infty$. That series then "operates" on $\epsilon_{(t+1)}$ so that's where all the lags of $\epsilon_{(t+1)}$ come from. Hopefully that clarifies what's going on there.
mark leeds
CC BY-SA 4.0
2023-08-02T06:08:35.083Z
2023-08-02T06:24:50.057Z
2023-08-02T06:24:50.057Z
mark leeds
<p>In Hansen's RBC Model, random technology variable follows the process
λ_(t+1)=γλ_t+ε_(t+1)
How do the assumptions of shocks ε_(t) being positive and bounded above with mean of 1-γ imply that the mean of λ_(t) is 1 and the output is non-negative?</p>

<macroeconomics><rbc>
Hansen's Basic Model (RBC)
S.G.
# Hansen's Basic Model (RBC)

(Asked by: username_0 on 2023-07-29)

In Hansen's RBC Model, random technology variable follows the process
λ\_(t+1)=γλ\_t+ε\_(t+1)
How do the assumptions of shocks ε\_(t) being positive and bounded above with mean of 1-γ imply that the mean of λ\_(t) is 1 and the output is non-negative?

**Tags:** macroeconomics, rbc

---

## 1 answers

### Answer by username_1 on 2023-08-02. Score: 0 | Accepted answer

That's an AR(1) for $\lambda(t)$ so $|y|$ has to be less than 1.0 for stationarity.

We have:

1. $\lambda\_{t+1} - y \lambda\_{t} = \epsilon\_{(t+1)}$

First, using the lag operator, $L$, we can simplify 1. :

2. $\lambda\_{t+1}(1 - y L) = \epsilon\_{(t+1)}$
3. $\lambda\_{t+1} = \frac{\epsilon\_{(t+1)}}{(1 - y L)}$

Assuming that $0 \lt y \lt 1.0$ ( not sure if I need this assumption but, if it's negative, it can't be converted to an infinite geometric series so it seems to me like the assumption is needed ), one can write the denominator as an infinite geometric series: This gives:

4. $\lambda\_{t+1} = \sum\_{i=0}^{\infty}{ y^{i} \times \epsilon\_{t+1-i}}$

where $0 \lt y \lt 1.0$.

Now take the expectation of both sides:

5. $E(\lambda\_{t+1}) = \sum\_{i=0}^{\infty} y^{i} \times E(\epsilon\_{t+1-i})$

But the expectation of the $\epsilon\_t$ is $(1-y)$ so 5) becomes:

6. $E(\lambda\_{t+1}) = (1 - y) \sum\_{i=0}^{\infty} y^{i} $

But the geometric series can be re-written as one term which is $\frac{1}{1-y}$ so we obtain:

7. $E(\lambda\_{t+1}) = \frac{(1-y)}{(1-y)} = 1.0$

So, the mean of $\lambda\_{t+1} = 1.0$ and $\epsilon\_{t}$ is always non-negative so 4) implies that $\lambda\_{t+1}$ is always non-negative.

> **Comments:**
> **username_0**: Could you please explain Equation (2)? I didn't get how you have factored out λ_t+1 from the last term - what does L denote? And what is the economic intuition of the denominator in Equation (4)?  
> **username_1**: $L$ is a lag operator with the definition that $L(x_{t+1}) = x_{t}$ so it lags the time, $t$ by one period.  It's pretty much the same thing as what z does in a z-transform if you are familar with that ?. I'm not a math person so maybe one of the mathematicians on this list can explain it more clearly or in more depth. Hamilton's "Time Series Analysis" also has an explanation.  
> **username_1**: Do you want me to change $y$ to $\rho$ everywhere because using $y$ to me is confusing. It's easy enough to change. Basically, 4) says that $\lambda$ is really just an exponentially smoothed function of the past shocks $\epsilon_t$. The lag on $\lambda_{t+1}$ on the right hand side is gone because the response ( the left hand side )  is only a function of all the past shocks that have happened upto and including time $(t+1)$  
> **username_1**: If you're into macro ( or rational expectations ) or planning on getting into macro or rational expectations, definitely you should familiarize yourself with the lag operator. Macro uses a lot of time series so the lag operator is crucial to understanding the literature. Maybe I should say: "attempting to understand the literature"  because the macro-literature is quite difficult for me. IMHO, Lucas is the clearest writer and maybe the only clear writer ? His "Studies in Business Cycle Theory" has the clearest explanations of RE that I know of but its old so totally RE oriented.  
> **username_1**: One thing that might look confusing is that ( 1 - y L) is in the denominator and I said eartlier that the lag operator lags it input by one. But there's no $t$ on the input $y$. so that seems weird. What's happening there is that the lag operator is actually working on the numerator, $\epsilon_t$. The $y$ is just a constant so nothing happens to that. So, if you had say $\frac{1}{1-p} $ where $p$ was positive and less than 1.0, that expression can be re-written as an infinite series: $1 + p + p^2 + \ldots + p^n$ all the way to $\infty$.  
> **username_1**: The above is analogous to what I'm doing in 4) with the only difference being that the infinite series is $1 + y L +  y^2 L^2 + \ldots + y^n L^{n} $  all the way to $\infty$. That series then "operates" on $\epsilon_{(t+1)}$ so that's where all the lags of $\epsilon_{(t+1)}$ come from. Hopefully that clarifies what's going on there.  

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