Hi,
I was hoping that someone would kindly help me with the last part of this question. I've found the general solution, however I have no idea how to approach the last part 'show that the limiting value..."
Many thanks!
$\displaystyle \text{Have: } $
$\displaystyle p_t = (1+r)p_{t-1} + k (q^D(p_{t-1}) - q^S(p_{t-1})), \ q^D(p) = 3 - p, \ q^S(p) = p \ p_0 = 1.$
$\displaystyle \text{Thus } p_t = (1+r)p_{t-1} + k ( \ (3 - p_{t-1}) - (p_{t-1}) \) = (1+r)p_{t-1} + k(3 - 2p_{t-1})$
$\displaystyle = (1+r)p_{t-1} + 3k - 2kp_{t-1} = 3k + (1+r - 2k)p_{t-1}.$
$\displaystyle \text{So your formula is } p_t = 3k + (1+r - 2k)p_{t-1}.$
$\displaystyle \text{Since that's not homogeneous, do the following trick to "divvy up" 3k equally to both terms:}$
$\displaystyle \text{If }(p_t+x) = (1+r - 2k)(p_{t-1}+x),$
$\displaystyle \text{then } p_t = [-x + (1+r - 2k)x] + (1+r - 2k)p_{t-1},$
$\displaystyle \text{so solve } -x + (1+r - 2k)x = 3k \text{ to get } x = \frac{3k}{r-2k} \text{ for } r \ne 2k.$
$\displaystyle \text{Let } a_t = p_t + x. \text{ Then } a_t = (1+r - 2k)a_{t-1}.$
$\displaystyle \text{That can be easily and explicitly solved for } a_t,$
$\displaystyle \text{ and so thus for } p_t = a_t - x.$
$\displaystyle \text{Also note that the } r = 2k \text{ case is } p_t = 3k + p_{t-1},$
$\displaystyle \text{which is both also explicitly solveable - and very divergent.}$
$\displaystyle \text{Maybe try it again from here?}$
$\displaystyle p(t)=\frac{3 k}{2 k-r}-\frac{(k+r) (-2 k+r+1)^t}{2 k-r}$
Convergence occurs for:
$\displaystyle |-2 k+r+1|\leq 1$
Simplify to obtain:
$\displaystyle r/2 < k\leq 1+r/2$
then for k=1+r/2:
$\displaystyle p(t)=\frac{3 (2+r)}{4}-\frac{1}{4} (-1)^t (2+3 r) = \left(2+\frac{3 r}{2}\right)$
and,
$\displaystyle \lim_{k\to \frac{r}{2}} \, p(t)\to 1+\frac{3 r t}{2}$
$\displaystyle \text{Assuming that solution, convergence fails at }-2k+r+1 = -1 \text$
$\displaystyle \text{when }k + r \ne 0.$
$\displaystyle p_t \text{ oscillates between } 1 \text{ and } \frac{4k+r}{2k - r} \text{ when } k = 1 + \frac{r}{2}.$