For a first order difference equation xt = axt-1+ b, given the initial value to the...

 

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For a first order difference equation xt = axt-1+ b, given the initial value to the solution is if a 1 and I -(-1)+1 (20- b -a -a = 2 = To+ bt. if a 1. Explain how the solution behaves in each of the following cases and draw a graph with an example. (a) 0 < a < 1, o > => 0. So, o is greater than the steady state valuer, which is strictly positive. 1-a (b) -1 < a < 0, x 0. So, zo is less than the steady state value, which is strictly positive. (c) a 1, b>0, To > 0. (d) a 1, b < 0, 0 > 0. (e) a 1, b=0, 10 > 0. For a first order difference equation xt = axt-1+ b, given the initial value to the solution is if a 1 and I -(-1)+1 (20- b -a -a = 2 = To+ bt. if a 1. Explain how the solution behaves in each of the following cases and draw a graph with an example. (a) 0 < a < 1, o > => 0. So, o is greater than the steady state valuer, which is strictly positive. 1-a (b) -1 < a < 0, x 0. So, zo is less than the steady state value, which is strictly positive. (c) a 1, b>0, To > 0. (d) a 1, b < 0, 0 > 0. (e) a 1, b=0, 10 > 0.