I am reading about difference equations (I have very little background in them ) in an economics book and it has an example where there are two statements neither which I have been able to derive. So, I repeat below what is in the book:

Suppose we have the following difference equation:

$y_{t-1} - \rho \times y_{t} = x_{t} $.

A) The book states the following solution to the difference equation above:

$y^{1}_{t} = \rho^{-1} \sum_{j=0}^{\infty} \rho^{-j} x_{t-j} $

B) The book states that there are multiple solutions because $ y^{1}_{t} + c \times \rho^{-t} $ is also a solution for any real c.

Can anyone derive A) and B) using either lag operators or the theory of difference equations. I was able to solve A) by brute force but I don't like the approach. I don't see B) at all. I looked in Goldberg's book which seems like the bible for difference equations but I am not familiar with the book so maybe I missed something. Thanks.

  • $\begingroup$ what range has $t$? I would suggest that you fix one $y_t$, e.g. $y_0$ and then see if other values for $y$ can be derived $\endgroup$ – supinf Jul 21 '17 at 10:38
  • $\begingroup$ @supinf: I can show A) using brute force which I think is what you're suggesting. But I think there must be an approach that uses the lag operator where L(y_{t} = y_{t-1} As far as the range of t, it's just the positive integers. I could have and maybe should have used $n$ instead of $t$. My apologies for any confusion due to using $t$. $\endgroup$ – mark leeds Jul 21 '17 at 11:06
  • $\begingroup$ Notice that in my original send, $\rho$ was mistakenly raised to $j$ instead of $-j$ so I just fixed it. my apologies. $\endgroup$ – mark leeds Jul 21 '17 at 11:12

For A, re-write the equation as $-(\rho -L)y_t=x_t$, (where $L$ is the lag operator). Solve for $y_t$ to get $$y_t =-\frac{1}{\rho - L}x_t$$and expand the fraction via long division.

B. Just substitute in the new proposed equation, $$ y^1_{t-1} +c\rho^{-(t-1)} -\rho(y^1_t +c\rho^{-t})=x_t$$ or $$y^1_{t-1}-\rho y^1_{t} + c\rho^{-t+1} -c\rho^{-t+1}=x_t.$$ And since $y^1_t$ is a solution, so is the new function.

  • $\begingroup$ beautiful. thanks. $\endgroup$ – mark leeds Jul 26 '17 at 19:42
  • $\begingroup$ There's a typo on the second to last line in that the 3rd term on the left hand side should be $-\rho(y_{t}^{1} + c\rho^{-t})$ but the editor wouldn't let me change it. not a big deal but if you can fix that for future readers, that would be great. thanks. $\endgroup$ – mark leeds Jul 26 '17 at 21:16
  • $\begingroup$ Done! sorry for the sign bust. $\endgroup$ – Trurl Jul 27 '17 at 12:41
  • $\begingroup$ forgot to upvote. $\endgroup$ – mark leeds Oct 13 '17 at 20:10

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