<p>He who formulated the exercise is wrong, and this is why I am posting a full answer to a homework question.
This is a classic example where manipulating recursive relations can lead to different representations that may appear "different" and with different properties.  </p>

<p>$$\Delta Y_t \equiv Y_{t}-Y_{t-1} =  \mu + Y_{t-1} + \varepsilon_{t} - \mu - Y_{t-2} - \varepsilon_{t-1}$$</p>

<p>$$\implies \Delta Y_t = \Delta Y_{t-1} - \Delta \varepsilon_{t} \tag{1}$$</p>

<p>At the same time </p>

<p>$$Y_t = \mu + Y_{t-1} + \varepsilon _t \implies \Delta Y_t = \mu +\varepsilon _t \tag{2}$$</p>

<p>The right hand side of Eqs. $(1)$ and $(2)$ represent the same process. </p>

<p>In any case, <em>neither</em> of these are $MA(1)$.</p>

<p>Moving forward, which one to choose?  </p>

<p>A soft adoption of Occam's razor indicates that eq. $(2)$ is the simplest. A bit more specifically, we note that the manipulation resulting in $(1)$ hasn't "saved" us from the existence of a unit root and non-stationarity in a clear way. </p>

<p>Both therefore suggest to adopt $(2)$, that says that  $\{\Delta Y_t\}_t$ is the sum of constant and a $WN$ process. </p>

CC BY-SA 4.0
2018-07-04T18:49:00.107Z
@markleeds Page 5-6 of the link you mention discusses a model with a deterministic trend.
Alecos Papadopoulos
CC BY-SA 4.0
2018-07-04T19:55:03.257Z
@alex papadopoulos: Sorry, Alex. It's page 3. My take on it is that if you difference the expectations at time (t+s) and time (t+s-1), then the first difference of these expectations will be  e_(t+s) - e_(t+s-1) which is an MA(1) with $\theta = -1$. That's the only thing I can think of as far as why the question asks to show it.  But I agree that it's pretty weird to look at it that way.
mark leeds
CC BY-SA 4.0
2018-07-05T02:39:05.147Z
Alecos. I looked more carefully and it does make slightly more sense than I first thought because you don't need to difference the expectations. Basically,  take the equation between "and" and "so that" on page 3. Then,  lag it by one period and subtract it from itself. The result you obtain is  $y_s - y_{s-1} = \epsilon_s - \epsilon_{s-1}$. Still it's a weird MA(1) because the two noise terms on the right hand side could be added and thought of as pure noise since the MA parameter is -1. Interesting anyway.
mark leeds
CC BY-SA 4.0
2018-07-05T03:01:28.423Z
@markleeds Mark, I don't get your calculations. I find 
$$y_{t+s-1} = y_0+b(t+s-1) +\sum_1^tu +\sum_{t+1}^{t+s-1}u$$

and so 

$$y_{t+s} - y_{t+s-1} = b+u_{t+s}$$  Once we have assumed a starting value, only the frontier terms are lagged
Alecos Papadopoulos
CC BY-SA 4.0
2018-07-05T05:38:45.840Z
You are absolutely correct. I made a mistake by doing it in my head. I learned from my mistakes and  proceeded to do it on paper to make sure no dumb mistakes and it turns out that the second difference is MA(1), so $y_{t+s} - y_{t+s-1} =  y_{t+s-1} - y_{t+s-2} + u_{t+s} - u_{t+s-1}$ so it can be viewed as an ARIMA(0,2,1). But that's not what the question asked so you are absoliutely correct that the question is wrong. Thanks for fix.
mark leeds
CC BY-SA 4.0
2018-07-05T05:48:01.793Z
Alecos: my apologies for noise also. I'll be more careful in the future. It's pretty bad though when a question asks you to do something that's not possible !!!!! I forget the book but I recently read an amazon review where the author purposely leaves  typos in the book ( some technical book, either econometrics or DSP. I don't remember ) to make sure that readers read carefully. that's definitely not a book I that  want to purchase !!!!!!! all the best and thanks again.
mark leeds
CC BY-SA 4.0
2018-07-05T08:02:00.523Z
@markleeds Mistakes on purpose is a cardinal sin in educational material. Conflicting results obtained by purposeful mistakes are good tools for clarifying issues though, as long a the writer asks the reader to find out what went wrong. See for example this post, https://economics.stackexchange.com/q/18811/61
Alecos Papadopoulos
CC BY-SA 4.0
2018-07-05T10:22:06.187Z
I agree but the author of the book isn't even hinting where the typos are. that's terrible.  I'll check  out the link.
mark leeds
CC BY-SA 4.0
2018-07-04T18:34:42.933Z
2018-07-04T18:34:42.933Z
Alecos Papadopoulos
<p>I have the following assignment to solve but I'm not sure if I solved it correctly.</p>

<p><strong>Questions</strong></p>

<blockquote>
  <p>Let the stochastic process $(Y_t)_t$ be defined by $Y_t = \mu + Y_{t-1} + \varepsilon _t$ with $(\varepsilon _t)_t\sim \mathrm{WN}(0,1)$.</p>
  
  <p>a) Calculate the expected value and the variance of $(\Delta Y_t)_t$.</p>
  
  <p>b) Prove that $(\Delta Y_t)_t\sim \mathrm{MA}(1)$ and calculate the autocovariance function of $(\Delta Y_t)_t$.</p>
</blockquote>

<p><strong>My solutions</strong></p>

<p>a) \begin{eqnarray}
Y_{t+1} &amp;=&amp; \mu + Y_t + \varepsilon_{t+1}\\[1ex]
\implies \Delta Y_t = Y_{t+1}-Y_t &amp;=&amp; \mu +\varepsilon_{t+1}\\[1ex]
\mathrm E(\Delta Y_t ) &amp;=&amp; \mu + \mathrm E(\varepsilon_{t+1}) = \mu \hspace{6cm} \\[1ex]
\mathrm {Var}(\Delta Y_t) &amp;=&amp; \mathrm{Var}(\varepsilon _{t+1}) = 1
\end{eqnarray}</p>

<p>b) \begin{eqnarray}
(\Delta Y_t)_t &amp;=&amp; \mu + \varepsilon_{t+1} + 0\cdot \varepsilon_t \\[1ex]
\implies \mathrm{ACV} &amp;=&amp; \mathrm{Cov}(\Delta Y_t, \Delta Y_{t-h})\\
 &amp;=&amp; \mathrm{Cov}(\varepsilon_{t+1}, \varepsilon_{t-h+1}) = \left\{\begin{array}{ll}
1 &amp; h=0 \\
0 &amp; \text{otherwise}
\end{array}\right.
\end{eqnarray}</p>

<p>What do you think?</p>

<p>
</p>

CC BY-SA 4.0
2018-06-29T08:29:36.227Z
Please read our [FAQ](https://economics.meta.stackexchange.com/questions/1465/economics-se-policy-on-homework-questions)!
"Do not merely post a scan or image of the whole question, nor of your attempted answer."
Giskard
CC BY-SA 4.0
2018-06-29T08:41:59.437Z
I'm sorry! I tried to do it but it wouldn't show my math formulas correctly.
rdvl0
CC BY-SA 4.0
2018-06-29T08:49:30.230Z
Hi: your acf is incorrect. write the stochastic process for $Y_{t-1}$ and then subtract it from the process for $Y_t$ and you will see that you end up with two epsilon terms which makes the acf at lag one non zero which implies that the process is MA(1). .
mark leeds
CC BY-SA 4.0
2018-06-30T00:05:00.110Z
@rdvl0 try this for fomatting your question https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference
EconJohn
CC BY-SA 4.0
2018-07-02T13:26:23.157Z
@denesp learning TeX isn't simple. Seeing as I think this question fits the 'requirements' of showing effort, I think the best thing for us to do is to edit the TeX ourselves. https://stackoverflow.blog/2018/04/26/stack-overflow-isnt-very-welcoming-its-time-for-that-to-change/
ahorn
CC BY-SA 4.0
2018-07-02T14:29:46.397Z
@markleeds $Y_t - Y_{t-1} = \mu + \varepsilon _t$. I'm not getting how this is MA(1).
ahorn
CC BY-SA 4.0
2018-07-02T15:11:25.277Z
@ahorn You are welcome to take part in the debate at the meta site and there you can try to change current consensus.
Giskard
CC BY-SA 4.0
2018-07-03T03:36:49.807Z
@ahorn: forgert my previous comment.. page 19 and 29 of the link below provides the properties of the process.  Based on that, it might be possible to show what you want to show. I'm gonna try it when I have more time. Right now, I don't see it. It's always possible that the thing you're being asked to show is not true. I hope link here helps. http://www.personal.psu.edu/asb17/old/sta4853/files/sta4853-2.pdf
mark leeds
CC BY-SA 4.0
2018-07-03T03:44:02.243Z
@ahorn: I only glanced but I'm pretty certain that page 5 of this shows it. http://www2.econ.iastate.edu/classes/econ672/falk/_notes/lecture_13_modeling_trends.pdf
mark leeds
CC BY-SA 4.0
2018-07-03T06:11:07.743Z
@rdvl0 see mark leeds's comments. I'm not going to bother (to put it bluntly ;) ).
ahorn
CC BY-SA 4.0
2018-07-03T12:31:34.467Z
@ahorn: I mistakenly thought that you had asked the question. my apologies.
mark leeds
CC BY-SA 4.0
2019-04-30T22:14:43.007Z
It seems that you should be able to give the covariance as simply $0$ for $h \neq0$. Including the value for $h=0$ is kinda redundant, since of course it's $1$. But of course your instructor's preferences take precedence.
Acccumulation
<mathematical-economics><statistics>
Did I calculate the expected value and the variance correctly?
rdvl0
# Did I calculate the expected value and the variance correctly?

(Asked by: username_0 on 2018-06-29)

I have the following assignment to solve but I'm not sure if I solved it correctly.

**Questions**

> Let the stochastic process $(Y\_t)\_t$ be defined by $Y\_t = \mu + Y\_{t-1} + \varepsilon \_t$ with $(\varepsilon \_t)\_t\sim \mathrm{WN}(0,1)$.
>
> a) Calculate the expected value and the variance of $(\Delta Y\_t)\_t$.
>
> b) Prove that $(\Delta Y\_t)\_t\sim \mathrm{MA}(1)$ and calculate the autocovariance function of $(\Delta Y\_t)\_t$.

**My solutions**

a) \begin{eqnarray}
Y\_{t+1} &=& \mu + Y\_t + \varepsilon\_{t+1}\\[1ex]
\implies \Delta Y\_t = Y\_{t+1}-Y\_t &=& \mu +\varepsilon\_{t+1}\\[1ex]
\mathrm E(\Delta Y\_t ) &=& \mu + \mathrm E(\varepsilon\_{t+1}) = \mu \hspace{6cm} \\[1ex]
\mathrm {Var}(\Delta Y\_t) &=& \mathrm{Var}(\varepsilon \_{t+1}) = 1
\end{eqnarray}

b) \begin{eqnarray}
(\Delta Y\_t)\_t &=& \mu + \varepsilon\_{t+1} + 0\cdot \varepsilon\_t \\[1ex]
\implies \mathrm{ACV} &=& \mathrm{Cov}(\Delta Y\_t, \Delta Y\_{t-h})\\
&=& \mathrm{Cov}(\varepsilon\_{t+1}, \varepsilon\_{t-h+1}) = \left\{\begin{array}{ll}
1 & h=0 \\
0 & \text{otherwise}
\end{array}\right.
\end{eqnarray}

What do you think?

**Tags:** mathematical-economics, statistics
> **Comments:**
> **username_1**: Please read our [FAQ](https://economics.meta.stackexchange.com/questions/1465/economics-se-policy-on-homework-questions)! "Do not merely post a scan or image of the whole question, nor of your attempted answer."  
> **username_0**: I'm sorry! I tried to do it but it wouldn't show my math formulas correctly.  
> **username_2**: Hi: your acf is incorrect. write the stochastic process for $Y_{t-1}$ and then subtract it from the process for $Y_t$ and you will see that you end up with two epsilon terms which makes the acf at lag one non zero which implies that the process is MA(1). .  
> **username_3**: @rdvl0 try this for fomatting your question https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference  
> **username_4**: @denesp learning TeX isn't simple. Seeing as I think this question fits the 'requirements' of showing effort, I think the best thing for us to do is to edit the TeX ourselves. https://stackoverflow.blog/2018/04/26/stack-overflow-isnt-very-welcoming-its-time-for-that-to-change/  
> **username_4**: @markleeds $Y_t - Y_{t-1} = \mu + \varepsilon _t$. I'm not getting how this is MA(1).  
> **username_1**: @ahorn You are welcome to take part in the debate at the meta site and there you can try to change current consensus.  
> **username_2**: @ahorn: forgert my previous comment.. page 19 and 29 of the link below provides the properties of the process.  Based on that, it might be possible to show what you want to show. I'm gonna try it when I have more time. Right now, I don't see it. It's always possible that the thing you're being asked to show is not true. I hope link here helps. http://www.personal.psu.edu/asb17/old/sta4853/files/sta4853-2.pdf  
> **username_2**: @ahorn: I only glanced but I'm pretty certain that page 5 of this shows it. http://www2.econ.iastate.edu/classes/econ672/falk/_notes/lecture_13_modeling_trends.pdf  
> **username_4**: @rdvl0 see mark leeds's comments. I'm not going to bother (to put it bluntly ;) ).  
> **username_2**: @ahorn: I mistakenly thought that you had asked the question. my apologies.  
> **username_5**: It seems that you should be able to give the covariance as simply $0$ for $h \neq0$. Including the value for $h=0$ is kinda redundant, since of course it's $1$. But of course your instructor's preferences take precedence.  

---

## 1 answers

### Answer by username_6 on 2018-07-04. Score: 1

He who formulated the exercise is wrong, and this is why I am posting a full answer to a homework question.
This is a classic example where manipulating recursive relations can lead to different representations that may appear "different" and with different properties.

$$\Delta Y\_t \equiv Y\_{t}-Y\_{t-1} = \mu + Y\_{t-1} + \varepsilon\_{t} - \mu - Y\_{t-2} - \varepsilon\_{t-1}$$

$$\implies \Delta Y\_t = \Delta Y\_{t-1} - \Delta \varepsilon\_{t} \tag{1}$$

At the same time

$$Y\_t = \mu + Y\_{t-1} + \varepsilon \_t \implies \Delta Y\_t = \mu +\varepsilon \_t \tag{2}$$

The right hand side of Eqs. $(1)$ and $(2)$ represent the same process.

In any case, *neither* of these are $MA(1)$.

Moving forward, which one to choose?

A soft adoption of Occam's razor indicates that eq. $(2)$ is the simplest. A bit more specifically, we note that the manipulation resulting in $(1)$ hasn't "saved" us from the existence of a unit root and non-stationarity in a clear way.

Both therefore suggest to adopt $(2)$, that says that $\{\Delta Y\_t\}\_t$ is the sum of constant and a $WN$ process.

> **Comments:**
> **username_6**: @markleeds Page 5-6 of the link you mention discusses a model with a deterministic trend.  
> **username_2**: @alex papadopoulos: Sorry, Alex. It's page 3. My take on it is that if you difference the expectations at time (t+s) and time (t+s-1), then the first difference of these expectations will be  e_(t+s) - e_(t+s-1) which is an MA(1) with $\theta = -1$. That's the only thing I can think of as far as why the question asks to show it.  But I agree that it's pretty weird to look at it that way.  
> **username_2**: Alecos. I looked more carefully and it does make slightly more sense than I first thought because you don't need to difference the expectations. Basically,  take the equation between "and" and "so that" on page 3. Then,  lag it by one period and subtract it from itself. The result you obtain is  $y_s - y_{s-1} = \epsilon_s - \epsilon_{s-1}$. Still it's a weird MA(1) because the two noise terms on the right hand side could be added and thought of as pure noise since the MA parameter is -1. Interesting anyway.  
> **username_6**: @markleeds Mark, I don't get your calculations. I find  $$y_{t+s-1} = y_0+b(t+s-1) +\sum_1^tu +\sum_{t+1}^{t+s-1}u$$  and so   $$y_{t+s} - y_{t+s-1} = b+u_{t+s}$$  Once we have assumed a starting value, only the frontier terms are lagged  
> **username_2**: You are absolutely correct. I made a mistake by doing it in my head. I learned from my mistakes and  proceeded to do it on paper to make sure no dumb mistakes and it turns out that the second difference is MA(1), so $y_{t+s} - y_{t+s-1} =  y_{t+s-1} - y_{t+s-2} + u_{t+s} - u_{t+s-1}$ so it can be viewed as an ARIMA(0,2,1). But that's not what the question asked so you are absoliutely correct that the question is wrong. Thanks for fix.  
> **username_2**: Alecos: my apologies for noise also. I'll be more careful in the future. It's pretty bad though when a question asks you to do something that's not possible !!!!! I forget the book but I recently read an amazon review where the author purposely leaves  typos in the book ( some technical book, either econometrics or DSP. I don't remember ) to make sure that readers read carefully. that's definitely not a book I that  want to purchase !!!!!!! all the best and thanks again.  
> **username_6**: @markleeds Mistakes on purpose is a cardinal sin in educational material. Conflicting results obtained by purposeful mistakes are good tools for clarifying issues though, as long a the writer asks the reader to find out what went wrong. See for example this post, https://economics.stackexchange.com/q/18811/61  
> **username_2**: I agree but the author of the book isn't even hinting where the typos are. that's terrible.  I'll check  out the link.  

---

