Help Understanding Sigma Notation

[Adrian_]

First off, I'm not 100% that I'm in the right forum so please excuse if not. Secondly if you can help me with this you are a great person and I thank you so very much

Working through a textbook solution but not following the process.

The textbook arrives at the following equation.

. . . . .\(\displaystyle \large{ y_t\, =\, \epsilon_t\, +\, 0.3\, \epsilon_{t - 1}\, +\, 0.8\, (0.3)\, \epsilon_{t -1}\, +\, 0.8^2\, (0.3)\, \epsilon_{t -3}\, +\, ... \, A\, (0.8)^t }\)

They then determine through another separate process (which we don't need to get into) that A in the equation above equals zero. So A=0
So after removing that term on the far right of the equation. They then represent the rest of the equation as follows

. . . . .\(\displaystyle \large{ \displaystyle y_t\, =\, \epsilon_t\,\, +\, 0.3\, \sum_{i\, =\, 0}^{t- 2}\, (0.8)^i\, \epsilon_{t - i - 1} }\)

The piece of this equation above that I do not understand is t-2 (on top of the sigma). Can someone explain why they've included the t-2? I do not see how the second equation I've pasted, represents the first one. I don't understand why in the second equation the upper bound of the summation range is t-2.

Thanks again

 

[Adrian_]

Help Interpreting Sigma Notation

Hi All,
I am following a long a textbook solution and am having trouble understanding sigma notation.

The text arrives at the following equation:

And in a separate piece of the problem (which we don't need to get into) it is made clear that A=0. So that the term on the far right drops out. With that far right term gone, the text now summarizes the remaining terms as follows:

The piece I do not understand is the t-2 on top of the sigma. Can someone explain to me how the t-2 in the second equation helps to denote the sum in the first equation. In the first equation, it looks like they written the sum to continue infinitely, isn't that what the "...." means in the first equation? Why are they including t-2 in the second equation?

Thanks very much for any help.

 

[Deleted member 4993]

First off, I'm not 100% that I'm in the right forum so please excuse if not. Secondly if you can help me with this you are a great person and I thank you so very much

Working through a textbook solution but not following the process.

The textbook arrives at the following equation.

. . . . .\(\displaystyle \large{ y_t\, =\, \epsilon_t\, +\, 0.3\, \epsilon_{t - 1}\, +\, 0.8\, (0.3)\, \epsilon_{t -1}\, +\, 0.8^2\, (0.3)\, \epsilon_{t -3}\, +\, ... \, A\, (0.8)^t }\)

They then determine through another separate process (which we don't need to get into) that A in the equation above equals zero. So A=0
So after removing that term on the far right of the equation. They then represent the rest of the equation as follows

. . . . .\(\displaystyle \large{ \displaystyle y_t\, =\, \epsilon_t\,\, +\, 0.3\, \sum_{i\, =\, 0}^{t- 2}\, (0.8)^i\, \epsilon_{t - i - 1} }\)

The piece of this equation above that I do not understand is t-2 (on top of the sigma). Can someone explain why they've included the t-2? I do not see how the second equation I've pasted, represents the first one. I don't understand why in the second equation the upper bound of the summation range is t-2.

Thanks again

You have:

\(\displaystyle \displaystyle{y_t = \epsilon_t + 0.3\epsilon_{t-1} + 0.3(0.8)\epsilon_{t-2}+ 0.3(0.8)^2\epsilon_{t-3} + 0.3(0.8)^3\epsilon_{t-4} + ... + 0.3(0.8)^n\epsilon_{t-n}...+ }\)

The last term of the series should have '1' as the subscript of 'ε'. The first term has 't-1' as subscript. So we utilize the fact that:

1 = (t-1) - (t-2)

 

[Adrian_]

Thank you very much Subhotosh .

I am not sure I understand. How do we know that the last term in the series will have 1 as a subcript of ε? Do you mean that the last term of the series would be (0.8)(0.3)(ε₁) ?

I thought the series just continues with ε_t-₁ , ε_t-_{2 ,} ε_t-3 infinitely. How does it end with (0.8)(0.3)(ε₁) ? The series looks to me to be moving backward in t not forwards

Thanks very much again