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Be $Y_t=X_t + \epsilon_{1,t}$, in which $X_t = X_{t-1} + \epsilon_{2,t}$ and $E[\epsilon_{1,t}\epsilon_{2,s}] = 0 \forall t,s$. How could I say why this process is related with a model on the form $(1-B)Y_t = \epsilon_t + \theta \epsilon_{t-1}$?

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in my work, I made:

$ Y_{t-1} = X_{t-1} + \epsilon_{1,t-1} = X_t - \epsilon_{2,t} + \epsilon_{1,t-1} = Y_t - \epsilon_{1,t} - \epsilon_{2,t} + \epsilon_{1, t-1}$

So,

$Y_t - Y_{t-1} = \epsilon_{1,t} + \epsilon_{2,t} - \epsilon_{1, t-1}$

Then,

$(1-B)Y_t = \epsilon_t - \epsilon_{t-1}$

Would my sense be correct?

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  • $\begingroup$ difference the first model on both sides and you'll get some process for $\triangle Y_t$. Then, show that the autocorrelations of that process are the same as those of the model $(1-B)Y_t = \epsilon_t + \theta \epsilon_{t-1}$. $\endgroup$ – mlofton Oct 16 at 2:35
  • $\begingroup$ I updated my work, would it be correct? $\endgroup$ – Tullio Oct 16 at 14:27
  • $\begingroup$ Hi: I don't like the very last equation but take the one right before it. ( which I like ). That's a differenced process with an non-zero autocorrelation at lag 1. The model with the $\theta$ is a differenced ARIMA process with a non-zero autocorrelation at lag 1. (an ARIMA(0,1,1). So, by the theory of ARIMA processes, which says that the autocorrelations of a stationary and invertible ARIMA process imply the model, then they are equivalent models. $\endgroup$ – mlofton Oct 16 at 19:16
  • $\begingroup$ not getting... Could you write the math? $\endgroup$ – Tullio Oct 17 at 1:32
  • $\begingroup$ Hi: The variance ( which is cov at lag zero) of the ARIMA(0,1,1) is $\sigma^2 + \theta^2 \sigma^2$. The lag one covariance is $\theta$. The cov of the ARIMA(0,1,1) is zero at all lags greater than 1. Now, do the same thing for the model you derived. You get the same thing in that there will be some variance and a covariance at lag 1 and zero covariance at lags greater than one This means that the models are equivalent. Do you have Andrew Harvey's blue book, "structural models and the kalman filter" ? or can you get your hands on it. There's a nice discussion of this in there. $\endgroup$ – mlofton Oct 18 at 3:33

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