5
$\begingroup$

Let $Z_t = c_1Z_{t-1} + c_2Z_{t-2} + ... + c_nZ_{t-n} + c\epsilon_t$

where $Z_t, \epsilon_t \sim \mathtt{N}(0,1)$ are iid variables and $Z_s \sim \mathtt{N}(0,1)$ for all $s$.

Given the values of $c_i$ for $i = 1 ...n$ is there a closed form formula for $c$?

We can derive that $c = \sqrt{1 - c_1^2}$ for the case $n = 1$ and that $c = \sqrt{1 - c_1^2 - c_2^2 -\frac{2c_1^2c_2}{1-c_2}}$ for case $n=2$.

So it seems like there should be a nice closed form formula for $c$. However I was unable to work through the algebra for $n = 3$.

But I think it might be that I don't know enough in this topic and perhaps this is a standard piece of work and someone has already worked it all out.

Improve this question
$\endgroup$
  • $\begingroup$ The part $Z_s \sim N(0,1)$ is misleading to me. I don't see why the variance of $Z_s$ is fixed to one. In a first order autoregressive model we have: $\hbox{Var}(Z_t) = c_1^2 \hbox{Var}(Z_{t-1}) + c^2 \hbox{Var}(\epsilon_t)$. Denoting the variance of $Z_t$ $\hbox{Var}(Z_t) = \sigma^2_Z$ and assuming a stationary process where the variance of $Z_t$ is the same as the variance of $Z_{t-1}$, we have $(1 - c_1^2)\sigma^2_Z = c^2\cdot 1$ and therefore $\sigma^2_Z = c^2/(1 - c_1^2)$, which is not equal to $1$ for any value of $c_1$ and $c$. $\endgroup$ – javlacalle Jul 11 '15 at 9:02
  • 1
    $\begingroup$ Are the values of the series $Z_t$, $t=1,2,...,n$ known? If so, can these values be part of the closed form solution you are looking for? $\endgroup$ – javlacalle Jul 11 '15 at 9:06
  • $\begingroup$ Denote $\gamma(h) = C[Z_t, Z_t+h]$. Then, because of the linearity of the covariance, $$\gamma(0) = C[Z_t, Z_t] = C[c_1 Z_{t-1} + \cdots, c_1 Z_{t-1} + \cdots] = c^2_1\gamma(1) + c^2_2\gamma(2) + \cdots c^2_p\gamma(p) + c^2.$$ That is, $$c^2 = \gamma(0) - \sum_{k=1}^p c^2_k\gamma(k).$$ So what you need is an expression for $\gamma$. There are a few different ways to compute it, see for example econweb.ucsd.edu/muendler/teach/00s/ps1-prt1.pdf $\endgroup$ – Hunaphu Jul 15 '15 at 9:38
3
+300
$\begingroup$

It's not strange you didn't calculate the AR(3) case. It's rather complicated! And no, there is no closed form for the AR(n)-case. For the AR(3) we start with the Yule-Walker equationsAR-model wikipedia(where $\gamma_j=\gamma_{-j}$):

  • $\gamma_1=c_1\gamma_0+c_2\gamma_{-1}+c_3\gamma_{-2}=c_1\gamma_0+c_2\gamma_{1}+c_3\gamma_{2}\\ \gamma_2=c_1\gamma_1+c_2\gamma_0+c_3\gamma_{-1}=c_1\gamma_1+c_2\gamma_0+c_3\gamma_{1}\\ \gamma_3=c_1\gamma_2+c_2\gamma_1+c_3\gamma_{0}$

    • Then we multiply the AR(3) expression by $Z_t$ and take the expectation: $ E[Z_t^2]=c_1E[Z_tZ_{t-1}]+c_2E[Z_tZ_{t-2}]+c_3E[Z_tZ_{t-3}]+c^2\Rightarrow \\ \gamma_0=c_1\gamma_1+c_2\gamma_2+c_3\gamma_3+c^2$

    • From Yule-Walker do we get: $ \gamma_1=\frac{c_1+c_2c_3}{1-c_2-c_1c_3-c_3^2}\gamma_0\\ \gamma_2=\left(\frac{c_1(c_1+c_2c_3)+c_3(c_1+c_2c_3)}{1-c_2-c_1c_3-c_3^2}+c_2\right)\gamma_0\\ \gamma_3=\left(\frac{(c_1^ 2+c_3c_1+c_2)(c_1+c_2c_3)}{1-c_2-c_1c_3-c_3^2}+c_2+c_3\right)\gamma_0$

    • Plugging these into the equation above (the second point) give what you want by setting $Var[Z_s]=\gamma_0=1$.

    • In Hamiltons book on time series (p59) he writes that the solutions for $\gamma_j$ takes the form: $\gamma_j=g_1\lambda_1^j+g_2\lambda_2^j+\cdots+g_p\lambda_p^j$

    where the eigenvalues are the solutions of $\lambda^p-c_1\lambda^{p-1}-c_2\lambda^{p-2}-\cdots-c_p=0$ (This is exactly the method outlined in the paper user Hunaphu has added). So my guess is that these equations will be very complicated for $p=5$ or above since no formula exists for the solution of an equation of degree five or higher. It has to be solved by e.g. elliptic functions or theta functions. So to find a general solution for the AR(n)-case you must find a general solution for the n'th-degree algebraic equation which does not exist.

Improve this answer
$\endgroup$
  • $\begingroup$ (+1) This is the idea that I had in mind when I asked the OP whether the data of $Z_t$ can be part of the closed form formula. This formula would then be given by your second point: $c = \sqrt{\gamma_0 + c_1\gamma_1 + c_2\gamma_2 + \cdots + c_n\gamma_n}$, where the terms $\gamma_i$ are the expressions of the sample autocovariances of order $i$ (and therefore depend on $Z_t$). $\endgroup$ – javlacalle Jul 14 '15 at 16:47
  • $\begingroup$ I don't see any reason to restrict the sample variance of $Z_t$ to 1, $\gamma_0=1$. In an AR process of order $n$ it will in general be different from zero. The OP did not clarify this issue. $\endgroup$ – javlacalle Jul 14 '15 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.