0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.