I first posted my question in the wrong forum (I did not know that there was an economic forum now) so I want to apologise for that by User Ben Bolker and User Robert. Also special thanks to User Robert who gave me the information with the econonomics forum!

So my question is about the Samuelson acceleration model. I am trying to help a friend of mine solve this question. I am not an economic studied person. I attached the question and my answers and conclusions. Maybe some of you can give me a little bit of help. In the first case, I did interpret the whole topic wrong and was confused because of the difference and thought this should be possible with an derivation (which was absolutely wrong, in my opionion now). I found an old analysis economics book and there were some interesting facts about the question (I also attached these facts).

The main question is, is it possible to bring the function back to the standard form of an inhomogen linear style with index inversion or something else. You can see what I have done attached, but I could not change the following from t-1 to t+1 because I did not know how.

Maybe some specialists here can help me.

enter image description here enter image description here enter image description here

Thank You!


  • $\begingroup$ Shouldn't the samuelson accelerator model be a second order difference equation because investment depends on consumption of the period before and therefore on income two periods before? $\endgroup$
    – PAS
    Mar 25 '18 at 13:43

There is nothing more to it than the equation

$$Y_t - (c_y+v)Y_{t-1} = C^a + I^a$$

This is a linear non-homogeneous first-order difference equation, and it is non-homogeneous because there is a non-zero constant, $C^a + I^a \neq 0$.

This is not some "habit in economics", but rather standard mathematical terminology.

  • $\begingroup$ Hello Alecos Papadopoulos, thanks for Your answer. So there is nothing to do after step 2 in my solution? Maybe only in step 3 showing that non-homogen function is not 0. Thank You for answering! Cheers $\endgroup$
    – 4lfr3d
    Oct 25 '17 at 11:34
  • $\begingroup$ @4lfr3d Nothing to do after arriving at the equation that appears also in my answer. $\endgroup$ Oct 25 '17 at 12:14

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