Mean and variance of difference operators on a time series process

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SUMMARY

The discussion focuses on computing the mean and variance of the second difference operator, denoted as ${\nabla}_{2}{Y}_{t}$, applied to a time series process defined as ${Y}_{t}={m}_{t}+{\varepsilon}_{t}$, where ${m}_{t}=a+bt$ and ${\varepsilon}_{t}$ is an i.i.d. sequence with mean 0 and variance ${\sigma}^{2}$. The user initially calculated the mean as 0 and the variance as $4{\sigma}^{2}$, but the correct results are a mean of $2b$ and a variance of $2{\sigma}^{2}$. The discrepancy arises from misunderstanding the application of the second difference operator and its effect on the mean and variance.

PREREQUISITES
  • Understanding of time series decomposition
  • Familiarity with the second difference operator in time series analysis
  • Knowledge of independent and identically distributed (i.i.d.) random variables
  • Basic statistical concepts of mean and variance
NEXT STEPS
  • Study the properties of the second difference operator in time series analysis
  • Learn about the impact of linear trends on time series mean and variance
  • Explore the derivation of mean and variance for different types of time series models
  • Investigate the role of i.i.d. sequences in statistical modeling
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Statisticians, data analysts, and researchers working with time series data who need to understand the implications of difference operators on mean and variance calculations.

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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