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For example, in this MA(2) model,

$y_t = u_t + \phi u_{t-2}$

$u_t$ is identically, independently, normally distributed with a mean of 0 and a variance of $\sigma^2$. (Does variance matter here?)

I think $E[u_{t}|y_{t-2}] = 0$ because it is impossible to represent $y_{t-2}$ using an equation containing $u_{t}$.

However, in the solution, it states that $E[u_{t-6}|y_{t-2}] = 0$ is also 0. But I think $y_{t-2}$ can be represented with an equation containing $u_{t-6}$ ($y_{t-2} = u_{t-2} + \phi (y_{t-4} - \phi u_{t-6})$). So why is $E[u_{t-6}|y_{t-2}] = 0$? Is it because "$u_t$ is identically, independently, normally distributed"? If yes, am I correct that any expected value of u conditional on y (no matter their index) is always 0 (including $E[u_t-2|y_{t}]$)?

Why is $E[y_{t-4}|y_{t-2}] = E[y_{t}|y_{t-2}]$? In the solution, it says "because $u_{t-2}|y_{t-2}$ and $u_{t-4}|y_{t-2}$ are independent conditionally". But how are the reasons given relevant? $y_{t}$ can not really be represented as equation containing $u_{t-2}$, $u_{t-4}$ and $y_{t-4}$.

Is it also analogous to autoregressive time series?

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From $y_t = u_t + \phi u_{t-2}$, we have

$$\left(\begin{matrix}u_{t-6}\\u_{t-4}\\u_{t-2}\\u_{t}\\y_{t-4}\\y_{t-2}\\y_{t} \end{matrix}\right) = \left(\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\phi&1&0&0\\0 & \phi&1&0&\\0& 0&\phi&1\end{matrix}\right) \left(\begin{matrix}u_{t-6}\\u_{t-4}\\u_{t-2}\\u_{t}\end{matrix}\right)$$

or

$$Y=Au$$ $u$ is multi-normal distributed, so $Y$ also follows multinormal distribution which determined by the mean vector and variance-covariance matrix.

$$\mathrm{E}(Y)=0$$ $$\mathrm{Var}(Y) = A\mathrm{Var}(u)A' = \sigma^2\left(\begin{matrix} I_{4\times 4}& \Sigma_{12}\\\Sigma_{12}'&\Sigma_{22}\end{matrix}\right)$$

where $$\Sigma_{22}=\left(\begin{matrix}\phi +1 &\phi & 0\\ \phi & \phi +1 &\phi \\ 0 &\phi & \phi +1\end{matrix}\right) $$ $$\Sigma_{12} = \left(\begin{matrix}\phi & 0 & 0 \\1 &\phi & 0\\ 0 & 1 &\phi \\ 0& 0 & \phi\end{matrix}\right) $$

Ignore the constant $\sigma^2$, because it appears in the numerator and denominator.

$\mathrm{E}(Y_{t-4}|Y_{t-2}) = E(Y_{t-4}) + \frac {Cov(Y_{t-4},Y_{t-2})}{Var(Y_{t-2})}(Y_{t-2}-E(Y_{t-2})) = 0 + \frac{\phi}{1+\phi}(Y_{t-2}-0) = \frac{\phi}{1+\phi}Y_{t-2}$

Replacing $Y_{t-4}$ by $Y_t$ the result is the same because $Cov(Y_{t-4},Y_{t-2}) = Cov(Y_{t},Y_{t-2})$.

For the rest, I think you can find the answers.

If not working with matrix, need following to get the variance and covariance based on $u_i$ is iid.

Let $Y_1 = a_1u_1+ ... + a_ku_k$ then $E(Y_1) = a_1E(u_1) +...+a_kE(u_k)$, $Var(Y_1) = a_1^2Var(u_1)+..+a_k^2Var(u_k)$

Let $Y_2 = b_1u_1+...+b_ku_k$ then $Cov(Y_1,Y_2)= a_1b_1Var(u_1)+...+a_kb_kVar(u_k)$

In your question about $Y_{t-2}$ and $Y_{t-4}$,

$\begin{align} Y_{t-2} = &1u_{t-2} + &\phi u_{t-4} + &0u_{t-6}\\ Y_{t-4} = &0u_{t-2} + &1 u_{t-4} + &\phi u_{t-6} \end{align}$

So $Cov(Y_{t-2}, Y_{t-4}) = \phi Var(u_{t-4}) = \sigma^2 \phi$

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  • $\begingroup$ I am not sure how is your answer relevant for conditional expectation. $\endgroup$
    – Aqqqq
    Dec 15, 2018 at 17:58
  • $\begingroup$ For example, here onlinecourses.science.psu.edu/stat505/node/43 has the descriptions on the conditional distribution of multinormal distribution. $\endgroup$
    – user158565
    Dec 15, 2018 at 18:00
  • $\begingroup$ I read these, but I am still not sure whether and why the statements in my answer is correct. Would you mind if you explain with an example? For example, can you tell me whether and why $E[y_{t-4}|y_{t-2}] = E[y_{t}|y_{t-2}]$ is correct? $\endgroup$
    – Aqqqq
    Dec 15, 2018 at 18:08
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    $\begingroup$ See the link above under "Bivariate Case" $\endgroup$
    – user158565
    Dec 15, 2018 at 19:08
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    $\begingroup$ Yes, no overlap ==> covariance = 0 ==> conditional expectation = unconditional expectation. $\endgroup$
    – user158565
    Dec 18, 2018 at 17:53

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