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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?

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1 Answer 1

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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

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

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

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

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

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  • $\begingroup$ Thank you for pointing me to RecurrenceTable! $\endgroup$
    – team-rf
    Commented Feb 17, 2016 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
    Commented Feb 19, 2016 at 10:11

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