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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?

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  • $\begingroup$ 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. $\endgroup$
    – Andrew M
    Jan 11, 2021 at 0:40
  • $\begingroup$ 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? $\endgroup$
    – Andrew M
    Jan 11, 2021 at 0:42
  • $\begingroup$ Hi Andrew, thanks for your feedback. I fixed the notation. I still don't quite really get how 4 is arrived at however. $\endgroup$
    – EB3112
    Jan 11, 2021 at 8:55
  • $\begingroup$ Is it $\mu_t$ or $\mu$? Notations are important... $\endgroup$
    – Bertrand
    Jan 11, 2021 at 9:54
  • $\begingroup$ $\mu_t$, apologies. $\endgroup$
    – EB3112
    Jan 11, 2021 at 10:01

1 Answer 1

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\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.

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  • $\begingroup$ Thank you DF, very much appreciated. $\endgroup$
    – EB3112
    Jan 11, 2021 at 13:21
  • $\begingroup$ @EB3112 if you think this answer answered your question consider accepting it $\endgroup$
    – 1muflon1
    Jan 11, 2021 at 17:03

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