3
$\begingroup$

I would like to illustrate rational expectations based on the following equation

p[[t+1]] == 1/(1 - ψ (β - 1)) m[[t+1]] - ψ (β - 1)/(1 - ψ (β - 1)) p[[t]]

by building a table or manipulating to make an algebraic form from t=1 to 10 while keeping p and m symbolic.

How could I go about it?

$\endgroup$
3
$\begingroup$

After @J.M., but note that you may have to define an initial value for m[0]:

FullSimplify@
  RecurrenceTable[{p[t + 1] == 
  1/(1 - ψ (β - 1)) m[t + 1] - ψ (β - 1)/(1 - ψ (β - 1)) p[t], 
   p[0] == 1}, p, {t, 1, 4}] // MatrixForm

$\frac{-\beta \psi +m(1)+\psi }{-\beta \psi +\psi +1}$

$\frac{(\beta -1)^2 \psi^2-(\beta -1) (m(1)+m(2)) \psi +m(2)}{((\beta -1) \psi -1)^2}$,

$\frac{(\beta -1)^3 \psi ^3-(\beta -1)^2 (m(1)+m(2)+m(3)) \psi ^2+(\beta -1) (m(2)+2 m(3)) \psi -m(3)}{((\beta -1) \psi -1)^3}$,

$\frac{(\beta -1)^4 \psi ^4-(\beta -1)^3 (m(1)+m(2)+m(3)+m(4)) \psi ^3+(\beta -1)^2 (m(2)+2 m(3)+3 m(4)) \psi ^2-(\beta -1) (m(3)+3 m(4)) \psi +m(4)}{(-\beta \psi +\psi +1)^4}$

$\endgroup$
  • $\begingroup$ Thank you for pointing me to RecurrenceTable! $\endgroup$ – team-rf Feb 17 '16 at 9:39
  • $\begingroup$ Consider $$ p_t=\frac{m_t}{\psi +1}+\frac{\psi \left(e_t p_{t+1}\right)}{\psi +1} $$ $$ e_t p_{t+1}=e_t \left(\frac{m_{t+1}}{\psi +1}+\frac{\psi \left(e_t p_{t+2}\right)}{\psi +1}\right)=\frac{e_t m_{t+1}}{\psi +1}+\frac{e_t \left(\psi \left(e_t p_{t+2}\right)\right)}{\psi +1}=\frac{e_t m_{t+2}}{\psi +1}+\frac{\psi \left(e_t p_{t+2}\right)}{\psi +1} $$ What I should get is $$ p_t=\frac{\psi \left(\frac{e_t m_{t+2}}{\psi +1}+\frac{\psi \left(e_t p_{t+2}\right)}{\psi +1}\right)}{\psi +1}+\frac{m_t}{\psi +1} $$ I'd like to expand to n cycles $\endgroup$ – team-rf Feb 19 '16 at 10:11

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.