<p>Hi: I don't know where you got the expression for $d_{t-1} - p_{t-1}$ but assuming that's true, then I think you are basically there. Take your final expression and  consider the two lagged white noise terms first. They can be added together to make one lagged white noise term because I'm pretty certain that the sum of two white noise processes is still white noise. Similarly, the two non lagged white noise terms can be added to make one non-lagged white noise term. So, you end up with an MA(1) so the whole expression is ARMA(2,1). Granted the MA(1) part will have some complicated constant coefficients on both the lagged and non-lagged term but they are not terribly important. Also, you can always normalize so that the non-lagged coefficient of the MA(1) piece is 1.0.  If you or anyone thinks this reasoning is flawed, let me know because I don't see the flaw in it and am all ears. Thanks.</p>

<p>$\boldsymbol{EDIT}$</p>

<p>Elbarto: I've dealt with ARIMA models quite a bit over the last 5-6 years so this was bothering me so I kept at it. It turns out ( after many readings that may have been in my sub-conscious initially ) that the 4 last terms of your original expression do represent an MA(1) process. The mistake I made is that my argument was el-wrongo. The following is the argument that's used in the literature-texts for showing what process any ARIMA type "process" reduces to. </p>

<p>1) Construct the autocorrelations at each lag of the process.</p>

<p>2) If that  calculation results in a functional form that behaves in the same way (in terms of its shape) as some well known ARIMA model, then the original unknown process is equivalent to that well known process. </p>

<p>Unfortunately, we were unable to derive the process was using your methodology which is more direct and, to me, more intuitive.  The 1) and 2) argument above stems from the fact that any stationary and invertible ARIMA model has a 1 to 1 mapping between the model parameters and the model  autocorrelations. Therefore, if you know the autocorrelation shape, you know the process. I'm not absolutely sure if the stationary part of the theorem is even needed and, in fact, I don't even know if it's an actual theorem. I just think of it as a theorem but it is true.  The best analogy to this argument would be knowing the impulse response in the DSP framework. Still, for the intuition,  I like your approach better.</p>

<p>So, to make a long story short,  using this approach,  take those 4 terms as some unknown  ARIMA process and calculate its autocorrelation. Since that process clearly has an non-zero autocorrelation at lag 1  and zero after that, it is an MA(1) process.  This is a specific case of the generic argument used in the literature  but it's slightly unsatisfying given how your method seems quite reasonable also. My thinking is that the  one to one mapping  theorem must be equivalent to what you (and I later) were trying to do. </p>

<p>Finally, if you want to be bothered ( I didn't want to be ), you could calculate the autocorrelation of that 4 term process. It will be some function of the three parameters. Then set that function equal to $\frac{\theta}{(1+\theta^2)}$ and  hopefully ( it looked ugly to me  and I'd probably make a mistake somewhere even if I could be bothered) solve for $\theta$. That implies that the four term  process is  an MA(1) with parameter $\theta$. Anyway, the usefulness of calculating it is that you would then have $\theta$ as some function of of the other parameters: $b_{x}$, $b_{y}$ and $\rho$.</p>

<p>With that in hand, it may become more clear as to what the algebraic argument is that would show the same thing ??? We must be thinking about it in some incorrect way that I can't see (a scaling-normalization issue perhaps ?).</p>

<p>I'm reasonably satisfied now so, if you can't be bothered, I understand. All the best and thanks for neat question. I learned something from you and also during my googling journey. </p>

<p>$\boldsymbol{Last~~ Comment}$</p>

<p>Elbarto: Just one last thing. Keep in mind the same autocorrleation argument could have been used for the full expression that includes the AR(2) term. I have no idea how difficult that is but it's not nearly as easy as the MA(1) because now the ACF dies out slowly or oscillates depending on the roots. In fact, I bet it would be a lot more work than you did to show the AR(2) directly so not really worth pursuing.</p>

CC BY-SA 4.0
2018-09-15T12:41:55.190Z
@elbarto: I see now where you got the expression for $d_{t-1} - p_{t-1}$ and I'll trust you on the campbell-shiller decomposition relation. Nice derivation.
mark leeds
CC BY-SA 4.0
2018-09-15T13:03:38.447Z
@Alberto: In O.D  Anderson's text, "Time Series Analysis and Forecasting", it is shown that the sum of two AR(1) processes is ARMA(2,1), ( I didn't read it carefully but there are some conditions that need to be met )  so that gives me more confidence that what you did is correct. All the best and thanks for interesting question.
mark leeds
CC BY-SA 4.0
2018-09-15T14:02:43.983Z
Thanks for that! I'll have a look at the reference you provided. I can see that I'm almost there but the hard part is defining the white noise terms. Say I let the MA(1) part be $\varepsilon_t = \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y}$, this doesn't quite work because then it means that $\varepsilon_{t-1} = \frac{\delta_{x, t-1}}{1-\rho b_x} + \frac{\delta_{y, t-1}}{1-\rho b_y}$ but the coefficients of $-b_x$ and $-b_y$ in the above ruins things... My gut feeling is that $q = 1$ but I can't seem to find what to define $\varepsilon_t$.
elbarto
CC BY-SA 4.0
2018-09-15T14:33:03.340Z
I couldn't seem to find the Anderson text you referenced, do you think you can lend some insights as to how it's proven? In my case, I essentially just have a linear combination of two AR(1)'s, e.g., $C_1x_t + C_2y_t$ where $C_1=\frac{1}{1-\rho b_x}$ and $C_2=\frac{1}{1-\rho b_y}$ are the constants.
elbarto
CC BY-SA 4.0
2018-09-15T14:35:40.247Z
Hi Elbarto: I don't know why they ruin things ? You can multiply appropriately to get a common denominator which results in a linear combination of two white noise terms. If these two white noise terms occur at the same time , say, time $t$ or $t-1$ etc, then their sum will also be white noise. White noise is an abstract concept ( derivative of brownian motion )  but, if you  assume that they are normal,  then the proof ( that a linear combination of two normals is normal ) is well known.
mark leeds
CC BY-SA 4.0
2018-09-15T14:37:24.830Z
I'll try to find a proof somewhere else because the anderson proof is not so clear.
mark leeds
CC BY-SA 4.0
2018-09-15T14:47:08.950Z
What I mean specifically is this: We need to show that $$- b_x \frac{\delta_{y, t-1}}{1-\rho b_y}- b_y \frac{\delta_{x, t-1}}{1-\rho b_x}+  \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y} = \varepsilon_t + \theta_1 \varepsilon_{t-1}$$ for some white noise $\varepsilon_t$ and appropriate constant $\theta_1$. How can I define this $\varepsilon_t$? If I let $\varepsilon_t = \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y}$, then I must have $b_x = b_y$ (so I can define $\theta_1 = -b_x = -b_y$).
elbarto
CC BY-SA 4.0
2018-09-15T16:13:39.423Z
Not sure if this proof is clear but, at a glance, it's clearer than  the proof in Anderson. ( see link below ). Also,  I think I see what your saying regarding $\epsilon_t$ and $\epsilon_{t-1}$ but let me look more carefully tonight or tomorrow. https://pdfs.semanticscholar.org/d3f9/8f511c0e4ffaf16cce79c7b5ff57c7f14b54.pdf
mark leeds
CC BY-SA 4.0
2018-09-16T06:03:40.323Z
Elbarto: I thought about it and played around and I think you're right. Since you don't have a sum of two AR(1) processes but rather a weighted sum of two AR(1) processes, I don't think that theorem in granger and morris is relevant. Still, it's an interesting problem so, if you come up with anything, let me know. Also, my apologies for noise.
mark leeds
CC BY-SA 4.0
2018-09-17T07:22:39.280Z
@Elbarto: I modified-corrected my answer by adding below it and I'm not sure if people are alerted to that so I am sending this comment just to make sure you get it.
mark leeds
CC BY-SA 4.0
2018-09-17T22:50:37.103Z
Thank you so much. I never knew about the mapping between autocorrelation and the process, after computing the autocorrelations indeed it matches with an MA(1) process. I have accepted your excellent answer.
elbarto
CC BY-SA 4.0
2018-09-18T02:59:32.923Z
Elbarto: you're welcome. As I said, I learned a lot also.  someday, if you can be bothered to show what was going wrong with what you ( we) were trying to do earlier, it's appreciated. Also,  I knew where the one to one mapping statement was, I would let you know but I don't. It was in my thesis (2000) but I still don't know where I got it. I definitely didn't come up with it. All the best.
mark leeds
CC BY-SA 4.0
2018-09-18T03:28:27.777Z
Elbarto: I've given up trying to find that uniqueness theoirem but I did notice that I made a mistake when calculating the autocorrelation of an MA(1) at lag 1. The denominator should be $1 + \theta^2$ rather than $1 + \theta$. I  hope that didn't mess up whatever you already did. If it did, then  my apologies. I made the correction in the answer.
mark leeds
CC BY-SA 4.0
2018-09-15T04:00:19.030Z
2018-09-18T03:29:24.437Z
2018-09-18T03:29:24.437Z
mark leeds
<p>Let the log dividend growth evolve according to $\Delta d_{t+1} = \epsilon_{d, t+1}$ where $\epsilon_{d, t+1}$ is just white noise. Let the log returns be $r_{t+1} = x_t + y_t + \epsilon_{r, t+1}$ where $x_t = b_x x_{t-1} + \delta_{x, t}$ and $y_t = b_y y_{t-1} + \delta_{y, t}$ and $\epsilon_{r, t+1}$, $\delta_{x, t}$, and $\delta_{y, t}$ are all white noise. Solve for the dividend price ratio, $d_t - p_t$ and show that it is an $ARMA(p, q)$ process, find $p$ and $q$.</p>

<p>What I did was to start off with the famous Campbell-Shiller decomposition: $d_t - p_t  = E_t \sum_{j=1}^{\infty} \rho^{j-1}(r_{t+j} - \Delta d_{t+j})$ and one can easily show that $E_t(r_{t+j}) = b_x^{j-1} x_t + b_y^{j-1} y_t$ and $E_t(\Delta d_{t+j}) = 0$ and so we get </p>

<p>$$d_t-p_t = \frac{x_t}{1-\rho b_x} + \frac{y_t}{1-\rho b_y} $$</p>

<p>I'm now stuck with trying to show that this is an ARMA(p, q) process. What I've tried is substituting in $x_t$ and $y_t$ to get:</p>

<p>$$d_t - p_t = b_x \frac{x_{t-1}}{1-\rho b_x} + b_y \frac{y_{t-1}}{1-\rho b_y}+ \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y}$$</p>

<p>But I am stuck on manipulating the RHS to form a $d_{t-1} - p_{t-1}$, specifically, I know that $d_{t-1}-p_{t-1} = \frac{x_{t-1}}{1-\rho b_x} + \frac{y_{t-1}}{1-\rho b_y}$ but the coefficients $b_x$ and $b_y$ on the RHS makes things difficult. </p>

<p>EDIT: I've done a bit more working out and here is what I've done:</p>

<p>\begin{align*}
d_t - p_t &amp; = \frac{x_t}{1-\rho b_x}  + \frac{y_t}{1-\rho b_y} \\
&amp; = \frac{b_x x_{t-1} + \delta_{x, t}}{1-\rho b_x} + \frac{b_y y_{t-1} + \delta_{y, t}}{1-\rho b_y} \\
&amp; = b_x \frac{x_{t-1}}{1-\rho b_x} + b_y \frac{y_{t-1}}{1-\rho b_y} +  \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y} \\
&amp; = b_x\left(d_{t-1} - p_{t-1} - \frac{y_{t-1}}{1-\rho b_y} \right) + b_y\left(d_{t-1} - p_{t-1} - \frac{x_{t-1}}{1-\rho b_x} \right)+  \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y} \\
&amp; = (b_x + b_y)(d_{t-1} - p_{t-1})-b_x\left(\frac{b_y y_{t-2} + \delta_{y, t-1}}{1-\rho b_y} \right)-b_y\left(\frac{b_x x_{t-2} + \delta_{x, t-1}}{1-\rho b_x} \right) +  \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y} \\
&amp; = (b_x + b_y)(d_{t-1} - p_{t-1}) - b_xb_y\left(\frac{x_{t-2}}{1-\rho b_x} + \frac{y_{t-2}}{1-\rho b_y} \right) - b_x \frac{\delta_{y, t-1}}{1-\rho b_y}- b_y \frac{\delta_{x, t-1}}{1-\rho b_x}+  \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y}  \\
&amp; = (b_x + b_y)(d_{t-1} - p_{t-1}) - b_xb_y(d_{t-2} - p_{t-2}) - b_x \frac{\delta_{y, t-1}}{1-\rho b_y}- b_y \frac{\delta_{x, t-1}}{1-\rho b_x}+  \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y}
\end{align*}</p>

<p>So it seems like there are two lags for $d_t-p_t$ but I am not sure how to manipulate the moving average terms...</p>

<econometrics><financial-economics><time-series><asset-pricing>
Show that the dividend price ratio is a ARMA(p, q) process
elbarto
# Show that the dividend price ratio is a ARMA(p, q) process

(Asked by: username_0 on 2018-09-14)

Let the log dividend growth evolve according to $\Delta d\_{t+1} = \epsilon\_{d, t+1}$ where $\epsilon\_{d, t+1}$ is just white noise. Let the log returns be $r\_{t+1} = x\_t + y\_t + \epsilon\_{r, t+1}$ where $x\_t = b\_x x\_{t-1} + \delta\_{x, t}$ and $y\_t = b\_y y\_{t-1} + \delta\_{y, t}$ and $\epsilon\_{r, t+1}$, $\delta\_{x, t}$, and $\delta\_{y, t}$ are all white noise. Solve for the dividend price ratio, $d\_t - p\_t$ and show that it is an $ARMA(p, q)$ process, find $p$ and $q$.

What I did was to start off with the famous Campbell-Shiller decomposition: $d\_t - p\_t = E\_t \sum\_{j=1}^{\infty} \rho^{j-1}(r\_{t+j} - \Delta d\_{t+j})$ and one can easily show that $E\_t(r\_{t+j}) = b\_x^{j-1} x\_t + b\_y^{j-1} y\_t$ and $E\_t(\Delta d\_{t+j}) = 0$ and so we get

$$d\_t-p\_t = \frac{x\_t}{1-\rho b\_x} + \frac{y\_t}{1-\rho b\_y} $$

I'm now stuck with trying to show that this is an ARMA(p, q) process. What I've tried is substituting in $x\_t$ and $y\_t$ to get:

$$d\_t - p\_t = b\_x \frac{x\_{t-1}}{1-\rho b\_x} + b\_y \frac{y\_{t-1}}{1-\rho b\_y}+ \frac{\delta\_{x, t}}{1-\rho b\_x} + \frac{\delta\_{y, t}}{1-\rho b\_y}$$

But I am stuck on manipulating the RHS to form a $d\_{t-1} - p\_{t-1}$, specifically, I know that $d\_{t-1}-p\_{t-1} = \frac{x\_{t-1}}{1-\rho b\_x} + \frac{y\_{t-1}}{1-\rho b\_y}$ but the coefficients $b\_x$ and $b\_y$ on the RHS makes things difficult.

EDIT: I've done a bit more working out and here is what I've done:

\begin{align\*}
d\_t - p\_t & = \frac{x\_t}{1-\rho b\_x} + \frac{y\_t}{1-\rho b\_y} \\
& = \frac{b\_x x\_{t-1} + \delta\_{x, t}}{1-\rho b\_x} + \frac{b\_y y\_{t-1} + \delta\_{y, t}}{1-\rho b\_y} \\
& = b\_x \frac{x\_{t-1}}{1-\rho b\_x} + b\_y \frac{y\_{t-1}}{1-\rho b\_y} + \frac{\delta\_{x, t}}{1-\rho b\_x} + \frac{\delta\_{y, t}}{1-\rho b\_y} \\
& = b\_x\left(d\_{t-1} - p\_{t-1} - \frac{y\_{t-1}}{1-\rho b\_y} \right) + b\_y\left(d\_{t-1} - p\_{t-1} - \frac{x\_{t-1}}{1-\rho b\_x} \right)+ \frac{\delta\_{x, t}}{1-\rho b\_x} + \frac{\delta\_{y, t}}{1-\rho b\_y} \\
& = (b\_x + b\_y)(d\_{t-1} - p\_{t-1})-b\_x\left(\frac{b\_y y\_{t-2} + \delta\_{y, t-1}}{1-\rho b\_y} \right)-b\_y\left(\frac{b\_x x\_{t-2} + \delta\_{x, t-1}}{1-\rho b\_x} \right) + \frac{\delta\_{x, t}}{1-\rho b\_x} + \frac{\delta\_{y, t}}{1-\rho b\_y} \\
& = (b\_x + b\_y)(d\_{t-1} - p\_{t-1}) - b\_xb\_y\left(\frac{x\_{t-2}}{1-\rho b\_x} + \frac{y\_{t-2}}{1-\rho b\_y} \right) - b\_x \frac{\delta\_{y, t-1}}{1-\rho b\_y}- b\_y \frac{\delta\_{x, t-1}}{1-\rho b\_x}+ \frac{\delta\_{x, t}}{1-\rho b\_x} + \frac{\delta\_{y, t}}{1-\rho b\_y} \\
& = (b\_x + b\_y)(d\_{t-1} - p\_{t-1}) - b\_xb\_y(d\_{t-2} - p\_{t-2}) - b\_x \frac{\delta\_{y, t-1}}{1-\rho b\_y}- b\_y \frac{\delta\_{x, t-1}}{1-\rho b\_x}+ \frac{\delta\_{x, t}}{1-\rho b\_x} + \frac{\delta\_{y, t}}{1-\rho b\_y}
\end{align\*}

So it seems like there are two lags for $d\_t-p\_t$ but I am not sure how to manipulate the moving average terms...

**Tags:** econometrics, financial-economics, time-series, asset-pricing

---

## 1 answers

### Answer by username_1 on 2018-09-15. Score: 2 | Accepted answer

Hi: I don't know where you got the expression for $d\_{t-1} - p\_{t-1}$ but assuming that's true, then I think you are basically there. Take your final expression and consider the two lagged white noise terms first. They can be added together to make one lagged white noise term because I'm pretty certain that the sum of two white noise processes is still white noise. Similarly, the two non lagged white noise terms can be added to make one non-lagged white noise term. So, you end up with an MA(1) so the whole expression is ARMA(2,1). Granted the MA(1) part will have some complicated constant coefficients on both the lagged and non-lagged term but they are not terribly important. Also, you can always normalize so that the non-lagged coefficient of the MA(1) piece is 1.0. If you or anyone thinks this reasoning is flawed, let me know because I don't see the flaw in it and am all ears. Thanks.

$\boldsymbol{EDIT}$

Elbarto: I've dealt with ARIMA models quite a bit over the last 5-6 years so this was bothering me so I kept at it. It turns out ( after many readings that may have been in my sub-conscious initially ) that the 4 last terms of your original expression do represent an MA(1) process. The mistake I made is that my argument was el-wrongo. The following is the argument that's used in the literature-texts for showing what process any ARIMA type "process" reduces to.

1) Construct the autocorrelations at each lag of the process.

2) If that calculation results in a functional form that behaves in the same way (in terms of its shape) as some well known ARIMA model, then the original unknown process is equivalent to that well known process.

Unfortunately, we were unable to derive the process was using your methodology which is more direct and, to me, more intuitive. The 1) and 2) argument above stems from the fact that any stationary and invertible ARIMA model has a 1 to 1 mapping between the model parameters and the model autocorrelations. Therefore, if you know the autocorrelation shape, you know the process. I'm not absolutely sure if the stationary part of the theorem is even needed and, in fact, I don't even know if it's an actual theorem. I just think of it as a theorem but it is true. The best analogy to this argument would be knowing the impulse response in the DSP framework. Still, for the intuition, I like your approach better.

So, to make a long story short, using this approach, take those 4 terms as some unknown ARIMA process and calculate its autocorrelation. Since that process clearly has an non-zero autocorrelation at lag 1 and zero after that, it is an MA(1) process. This is a specific case of the generic argument used in the literature but it's slightly unsatisfying given how your method seems quite reasonable also. My thinking is that the one to one mapping theorem must be equivalent to what you (and I later) were trying to do.

Finally, if you want to be bothered ( I didn't want to be ), you could calculate the autocorrelation of that 4 term process. It will be some function of the three parameters. Then set that function equal to $\frac{\theta}{(1+\theta^2)}$ and hopefully ( it looked ugly to me and I'd probably make a mistake somewhere even if I could be bothered) solve for $\theta$. That implies that the four term process is an MA(1) with parameter $\theta$. Anyway, the usefulness of calculating it is that you would then have $\theta$ as some function of of the other parameters: $b\_{x}$, $b\_{y}$ and $\rho$.

With that in hand, it may become more clear as to what the algebraic argument is that would show the same thing ??? We must be thinking about it in some incorrect way that I can't see (a scaling-normalization issue perhaps ?).

I'm reasonably satisfied now so, if you can't be bothered, I understand. All the best and thanks for neat question. I learned something from you and also during my googling journey.

$\boldsymbol{Last~~ Comment}$

Elbarto: Just one last thing. Keep in mind the same autocorrleation argument could have been used for the full expression that includes the AR(2) term. I have no idea how difficult that is but it's not nearly as easy as the MA(1) because now the ACF dies out slowly or oscillates depending on the roots. In fact, I bet it would be a lot more work than you did to show the AR(2) directly so not really worth pursuing.

> **Comments:**
> **username_1**: @elbarto: I see now where you got the expression for $d_{t-1} - p_{t-1}$ and I'll trust you on the campbell-shiller decomposition relation. Nice derivation.  
> **username_1**: @Alberto: In O.D  Anderson's text, "Time Series Analysis and Forecasting", it is shown that the sum of two AR(1) processes is ARMA(2,1), ( I didn't read it carefully but there are some conditions that need to be met )  so that gives me more confidence that what you did is correct. All the best and thanks for interesting question.  
> **username_0**: Thanks for that! I'll have a look at the reference you provided. I can see that I'm almost there but the hard part is defining the white noise terms. Say I let the MA(1) part be $\varepsilon_t = \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y}$, this doesn't quite work because then it means that $\varepsilon_{t-1} = \frac{\delta_{x, t-1}}{1-\rho b_x} + \frac{\delta_{y, t-1}}{1-\rho b_y}$ but the coefficients of $-b_x$ and $-b_y$ in the above ruins things... My gut feeling is that $q = 1$ but I can't seem to find what to define $\varepsilon_t$.  
> **username_0**: I couldn't seem to find the Anderson text you referenced, do you think you can lend some insights as to how it's proven? In my case, I essentially just have a linear combination of two AR(1)'s, e.g., $C_1x_t + C_2y_t$ where $C_1=\frac{1}{1-\rho b_x}$ and $C_2=\frac{1}{1-\rho b_y}$ are the constants.  
> **username_1**: Hi Elbarto: I don't know why they ruin things ? You can multiply appropriately to get a common denominator which results in a linear combination of two white noise terms. If these two white noise terms occur at the same time , say, time $t$ or $t-1$ etc, then their sum will also be white noise. White noise is an abstract concept ( derivative of brownian motion )  but, if you  assume that they are normal,  then the proof ( that a linear combination of two normals is normal ) is well known.  
> **username_1**: I'll try to find a proof somewhere else because the anderson proof is not so clear.  
> **username_0**: What I mean specifically is this: We need to show that $$- b_x \frac{\delta_{y, t-1}}{1-\rho b_y}- b_y \frac{\delta_{x, t-1}}{1-\rho b_x}+  \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y} = \varepsilon_t + \theta_1 \varepsilon_{t-1}$$ for some white noise $\varepsilon_t$ and appropriate constant $\theta_1$. How can I define this $\varepsilon_t$? If I let $\varepsilon_t = \frac{\delta_{x, t}}{1-\rho b_x} + \frac{\delta_{y, t}}{1-\rho b_y}$, then I must have $b_x = b_y$ (so I can define $\theta_1 = -b_x = -b_y$).  
> **username_1**: Not sure if this proof is clear but, at a glance, it's clearer than  the proof in Anderson. ( see link below ). Also,  I think I see what your saying regarding $\epsilon_t$ and $\epsilon_{t-1}$ but let me look more carefully tonight or tomorrow. https://pdfs.semanticscholar.org/d3f9/8f511c0e4ffaf16cce79c7b5ff57c7f14b54.pdf  
> **username_1**: Elbarto: I thought about it and played around and I think you're right. Since you don't have a sum of two AR(1) processes but rather a weighted sum of two AR(1) processes, I don't think that theorem in granger and morris is relevant. Still, it's an interesting problem so, if you come up with anything, let me know. Also, my apologies for noise.  
> **username_1**: @Elbarto: I modified-corrected my answer by adding below it and I'm not sure if people are alerted to that so I am sending this comment just to make sure you get it.  
> **username_0**: Thank you so much. I never knew about the mapping between autocorrelation and the process, after computing the autocorrelations indeed it matches with an MA(1) process. I have accepted your excellent answer.  
> **username_1**: Elbarto: you're welcome. As I said, I learned a lot also.  someday, if you can be bothered to show what was going wrong with what you ( we) were trying to do earlier, it's appreciated. Also,  I knew where the one to one mapping statement was, I would let you know but I don't. It was in my thesis (2000) but I still don't know where I got it. I definitely didn't come up with it. All the best.  
> **username_1**: Elbarto: I've given up trying to find that uniqueness theoirem but I did notice that I made a mistake when calculating the autocorrelation of an MA(1) at lag 1. The denominator should be $1 + \theta^2$ rather than $1 + \theta$. I  hope that didn't mess up whatever you already did. If it did, then  my apologies. I made the correction in the answer.  

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