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edit approved Jul 15, 2018 at 12:37
Rodrigo de Azevedo
edit approved Jul 15, 2018 at 12:37
Rodrigo de Azevedo
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general General solution for first order-order difference equation

I have the following first-order difference equation

$$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$$

where $L$ denotes the backshift operator, i.e., $L(x_{t}) = x_{t-1}$. I can obtain a solution to the followingthis difference equation heuristically, but I wasam wondering if there is a general procedure. The equation is:

$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$, where $L$ denotes the backshift operator so that $L(x_{t}) = x_{t-1}$.

A solution (or the, I'm not sure )the solution?) is $y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$.

but I expected there is a general-procedure formula.$$y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$$

I looked in in Goldberger's text and couldn't find it there. Any reference is appreciated also. Thanks.

general solution for first order difference equation

I can obtain a solution to the following difference equation heuristically, but I was wondering if there is a procedure. The equation is:

$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$, where $L$ denotes the backshift operator so that $L(x_{t}) = x_{t-1}$.

A (or the, I'm not sure ) solution is $y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$.

but I expected there is a general-procedure formula. I looked in Goldberger's text and couldn't find it there. Any reference is appreciated also. Thanks.

General solution for first-order difference equation

I have the following first-order difference equation

$$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$$

where $L$ denotes the backshift operator, i.e., $L(x_{t}) = x_{t-1}$. I can obtain a solution to this difference equation heuristically, but I am wondering if there is a general procedure. A solution (the solution?) is

$$y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$$

I looked in Goldberger's text and couldn't find it there. Any reference is appreciated also. Thanks.

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edit approved Jul 15, 2018 at 8:25
Amir Sagiv
edit approved Jul 15, 2018 at 8:25
Amir Sagiv
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Hi: I can obtain a solutonsolution to the following difference equation heuristically, but I was wondering if there is a procedure. The equation is:

$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$, where $L$ denotes the backshift operator so that $L(x_{t}) = x_{t-1}$.

A ( oror the  , I'm not sure ) solution is $y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$.

but I expected there to beis a general procedure-formulaprocedure formula. I looked in Goldberger's text and couldn't find it there either. Any reference is appreciated also. Thanks.

Hi: I can obtain a soluton to the following difference equation heuristically but I was wondering if there is a procedure. The equation is:

$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$ where $L$ denotes the backshift operator so that $L(x_{t}) = x_{t-1}$.

A ( or the  , I'm not sure ) solution is $y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$

but I expected there to be a general procedure-formula. I looked in Goldberger's text and couldn't find it there either. Any reference is appreciated also. Thanks.

I can obtain a solution to the following difference equation heuristically, but I was wondering if there is a procedure. The equation is:

$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$, where $L$ denotes the backshift operator so that $L(x_{t}) = x_{t-1}$.

A (or the, I'm not sure ) solution is $y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$.

but I expected there is a general-procedure formula. I looked in Goldberger's text and couldn't find it there. Any reference is appreciated also. Thanks.

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mark leeds
mark leeds
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general solution for first order difference equation

Hi: I can obtain a soluton to the following difference equation heuristically but I was wondering if there is a procedure. The equation is:

$y_{t} = \frac{x_{t}}{1-\rho L} + \epsilon_{t}$ where $L$ denotes the backshift operator so that $L(x_{t}) = x_{t-1}$.

A ( or the , I'm not sure ) solution is $y_{t} = \sum_{i = 0}^{\infty} \rho^{i}X_{t-i} + \epsilon_{t}$

but I expected there to be a general procedure-formula. I looked in Goldberger's text and couldn't find it there either. Any reference is appreciated also. Thanks.