<p><span class="math-container">\begin{align}y_t &amp;= \alpha + \theta_1y_{t-1}+u_t \\
&amp;= \alpha+\theta_1(\alpha + \theta_1y_{t-2}+u_{t-1}) + u_{t} \\
&amp;= (1+\theta_1) \alpha + \theta_1^2y_{t-2} + \theta_1u_{t-1}+u_{t} \\
&amp;=  (1+\theta_1) \alpha + \theta_1^2(\alpha + \theta_1y_{t-3}+u_{t-3}) + \theta_1u_{t-1}+u_{t} \\
&amp;=  (1+\theta_1 + \theta_1^2) \alpha + \theta_1^3y_{t-3} + \theta_1^3u_{t-3}+\theta_1u_{t-1}+u_{t} \\
&amp;= \dots \\
&amp;= (1+\theta_1+\dots+\theta_1^{t-1})\alpha+\theta_1^ty_0+\theta_1^{t}u_{0}+\dots+\theta_1u_{t-1}+u_{t} \\
&amp;= \mu_t+\theta_1^{t}u_{0}+\dots+\theta_1u_{t-1}+u_{t}
\end{align}</span></p>
<p>Hence</p>
<p><span class="math-container">$$E[y_t] = E[\mu_t]+\theta_1^{t}E[u_{0}]+\dots+\theta_1E[u_{t-1}]+E[u_{t}] = E[\mu_t]$$</span> since <span class="math-container">$E[u_{t}] = 0\ \forall\ t$</span></p>
<p>Finally, <span class="math-container">$$E[\mu_t] = (1+\theta_1+\dots+\theta_1^{t-1})\alpha+\theta_1^tE[y_0] = (1+\theta_1+\dots+\theta_1^{t-1})\alpha+\theta_1^ty_0 = \mu_t$$</span> where we assume that <span class="math-container">$y_0$</span> is non-random.</p>

CC BY-SA 4.0
2021-01-11T13:21:24.237Z
Thank you DF, very much appreciated.
EB3112
CC BY-SA 4.0
2021-01-11T17:03:40.330Z
@EB3112 if you think this answer answered your question consider accepting it
1muflon1
CC BY-SA 4.0
2021-01-11T11:10:11.307Z
2021-01-11T14:23:01.207Z
2021-01-11T14:23:01.207Z
D F
<p>Would appreciate some guidance on a matter of recursive substitution, where we have the AR model:</p>
<p><span class="math-container">$$y_t = \alpha +\theta_1y_{t-1}+ u_t$$</span></p>
<p>And</p>
<p><span class="math-container">$$E(y_t)= \mu_t$$</span></p>
<p>Where:</p>
<p><span class="math-container">$$\mu_t = (1+\theta_1 + \theta_1^2+..+\theta^{t-1})\alpha+\theta^ty_0$$</span></p>
<p>It follows by recursive substitution we get:</p>
<p><span class="math-container">$$y_t = \mu_t +(u_t +\theta_1u_{t-1}+\theta^2u_{t-2}+...+\theta^{t-1}u_1)$$</span></p>
<p>And subsequently:</p>
<p><span class="math-container">$$E[y_t] = E[\mu_t]+ E[(u_t +\theta_1u_{t-1}+\theta^2u_{t-2}+...+\theta^{t-1}u_1)]= \mu_t$$</span></p>
<p>Would someone be able to explain how the step concerning recursive substitution is obtained, moving from line 2,3,4?</p>

CC BY-SA 4.0
2021-01-11T00:40:50.993Z
the change in notation in equation 4 and 5 is confusing. Shouldn't equation 1 be $\epsilon_t$? Then rather than using $u_{t-j}$ in equation 4 and 5 it should be $\epsilon_{t-j}$. You are also missing $\mu$ in the third equality of equation 5.
Andrew M
CC BY-SA 4.0
2021-01-11T00:42:31.790Z
now if you are still confused once you fix the notation. Try plugging in the value for $y_{t-1}$ in equation 1 and keep going back in time. What patterns do you notice? How do they relate to equations 3 and 4?
Andrew M
CC BY-SA 4.0
2021-01-11T08:55:42.800Z
Hi Andrew, thanks for your feedback. I fixed the notation. I still don't quite really get how 4 is arrived at however.
EB3112
CC BY-SA 4.0
2021-01-11T09:54:10.370Z
Is it $\mu_t$ or $\mu$? Notations are important...
Bertrand
CC BY-SA 4.0
2021-01-11T10:01:35.083Z
$\mu_t$, apologies.
EB3112
<mathematical-economics><time-series>
Recursive Substitution in Time Series
EB3112
# Recursive Substitution in Time Series

(Asked by: username_0 on 2021-01-10)

Would appreciate some guidance on a matter of recursive substitution, where we have the AR model:

$$y\_t = \alpha +\theta\_1y\_{t-1}+ u\_t$$

And

$$E(y\_t)= \mu\_t$$

Where:

$$\mu\_t = (1+\theta\_1 + \theta\_1^2+..+\theta^{t-1})\alpha+\theta^ty\_0$$

It follows by recursive substitution we get:

$$y\_t = \mu\_t +(u\_t +\theta\_1u\_{t-1}+\theta^2u\_{t-2}+...+\theta^{t-1}u\_1)$$

And subsequently:

$$E[y\_t] = E[\mu\_t]+ E[(u\_t +\theta\_1u\_{t-1}+\theta^2u\_{t-2}+...+\theta^{t-1}u\_1)]= \mu\_t$$

Would someone be able to explain how the step concerning recursive substitution is obtained, moving from line 2,3,4?

**Tags:** mathematical-economics, time-series
> **Comments:**
> **username_1**: the change in notation in equation 4 and 5 is confusing. Shouldn't equation 1 be $\epsilon_t$? Then rather than using $u_{t-j}$ in equation 4 and 5 it should be $\epsilon_{t-j}$. You are also missing $\mu$ in the third equality of equation 5.  
> **username_1**: now if you are still confused once you fix the notation. Try plugging in the value for $y_{t-1}$ in equation 1 and keep going back in time. What patterns do you notice? How do they relate to equations 3 and 4?  
> **username_0**: Hi Andrew, thanks for your feedback. I fixed the notation. I still don't quite really get how 4 is arrived at however.  
> **username_2**: Is it $\mu_t$ or $\mu$? Notations are important...  
> **username_0**: $\mu_t$, apologies.  

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## 1 answers

### Answer by username_3 on 2021-01-11. Score: 4 | Accepted answer

\begin{align}y\_t &= \alpha + \theta\_1y\_{t-1}+u\_t \\
&= \alpha+\theta\_1(\alpha + \theta\_1y\_{t-2}+u\_{t-1}) + u\_{t} \\
&= (1+\theta\_1) \alpha + \theta\_1^2y\_{t-2} + \theta\_1u\_{t-1}+u\_{t} \\
&= (1+\theta\_1) \alpha + \theta\_1^2(\alpha + \theta\_1y\_{t-3}+u\_{t-3}) + \theta\_1u\_{t-1}+u\_{t} \\
&= (1+\theta\_1 + \theta\_1^2) \alpha + \theta\_1^3y\_{t-3} + \theta\_1^3u\_{t-3}+\theta\_1u\_{t-1}+u\_{t} \\
&= \dots \\
&= (1+\theta\_1+\dots+\theta\_1^{t-1})\alpha+\theta\_1^ty\_0+\theta\_1^{t}u\_{0}+\dots+\theta\_1u\_{t-1}+u\_{t} \\
&= \mu\_t+\theta\_1^{t}u\_{0}+\dots+\theta\_1u\_{t-1}+u\_{t}
\end{align}

Hence

$$E[y\_t] = E[\mu\_t]+\theta\_1^{t}E[u\_{0}]+\dots+\theta\_1E[u\_{t-1}]+E[u\_{t}] = E[\mu\_t]$$ since $E[u\_{t}] = 0\ \forall\ t$

Finally, $$E[\mu\_t] = (1+\theta\_1+\dots+\theta\_1^{t-1})\alpha+\theta\_1^tE[y\_0] = (1+\theta\_1+\dots+\theta\_1^{t-1})\alpha+\theta\_1^ty\_0 = \mu\_t$$ where we assume that $y\_0$ is non-random.

> **Comments:**
> **username_0**: Thank you DF, very much appreciated.  
> **username_4**: @EB3112 if you think this answer answered your question consider accepting it  

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