Mean and variance of difference operators on a time series process

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Discussion Overview

The discussion revolves around the mean and variance of the difference operator applied to a time series process, specifically focusing on the decomposition of a time series into a deterministic trend and a stochastic component. Participants are examining the calculations related to the second difference of the time series.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a time series decomposition where \(Y_t = m_t + \varepsilon_t\) with \(m_t = a + bt\) and computes the second difference \(\nabla_2 Y_t\) as \(Y_t - 2Y_{t-1} + Y_{t-2}\), leading to \(\varepsilon_t - 2\varepsilon_{t-1} + \varepsilon_{t-2}\).
  • This participant claims the mean is 0 and the variance is \(4\sigma^2\), but questions the provided answer of mean \(2b\) and variance \(2\sigma^2\).
  • Another participant asks whether the answer provided included detailed computations.
  • A subsequent reply confirms that the answer did not detail computations and suggests that the provided answer could be incorrect, seeking clarification on potential errors in the initial calculations.
  • Another participant speculates that the term \(2b\) for the variance might arise if \(m_t\) were defined as \(a + bt^2\), expressing confusion about how the \(2\sigma^2\) term is derived.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correctness of the calculations and the provided answer, indicating that multiple competing views remain without a consensus on the correct mean and variance.

Contextual Notes

There are unresolved questions about the definitions used for the time series components and the calculations of mean and variance, which may depend on the specific formulation of \(m_t\).

deba123
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$$\text{Consider the following decomposition of the time series }{Y}_{t}\text{ where }{Y}_{t}={m}_{t}+{\varepsilon}_{t},\text{ where }{\varepsilon}_{t}\text{ is a sequence of i.i.d }\left(0,{\sigma}^{2}\right)\text{ process. Compute the mean and variance of the process }{\nabla}_{2}{Y}_{t}\text{ when }:{m}_{t}=a+bt.$$
 
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Welcome to MHB! :D

Can you show what you have tried so far so our helpers have some idea where you are stuck and can then offer better help?
 
MarkFL said:
Welcome to MHB! :D

Can you show what you have tried so far so our helpers have some idea where you are stuck and can then offer better help?

$$O.K\: so\: here's\: what\: i\: did:
{\nabla}_{2}{Y}_{t}=\nabla(\nabla{Y}_{t})=\nabla(({Y}_{t})-({Y}_{t-1}))={Y}_{t}-2{Y}_{t-1}+{Y}_{t-2}.
Which\: after\: expanding\: comes\: out:\:
{\varepsilon}_{t}-2{\varepsilon}_{t-1}+{\varepsilon}_{t-2}.
So\: mean\: is\: 0 \:and\: variance\: is\: 4{\sigma}^{2}.$$
$$But\:the\:answer\: is :mean=2b\:and\: variance=2{\sigma}^{2}.$$ I don't get how the answer came about something like that. Please help.
 
Did the answer detail the computations?
 
girdav said:
Did the answer detail the computations?

No it didn't. But the answer could be wrong. I just want to know where I was wrong (or right) and may be if I was using the wrong definition. Thanks for your help.
 
The term $2b$ for the variance could be explained if we had considered $m_t=a+bt^2$, but I fail to see how the $2\sigma^2$ terms come from.
 

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