Mean and variance of difference operators on a time series process

[deba123]

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

 

[MarkFL]

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?

 

[deba123]

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.

 

[tarantella]

Did the answer detail the computations?

 

[deba123]

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.

 

[tarantella]

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.