Difference Equations
First Order Difference Equations
Following equation is the linear first order difference equation:
yt=ϕyt−1+wt.
Recursive Solution
Consider the following system
Date | Equation |
---|
0 | y0=ϕy−1+w0 |
1 | y1=ϕy0+w1 |
2 | y2=ϕy1+w2 |
⋮ | ⋮ |
t | yt=ϕyt−1+wt |
Given the initial conditions y−1 and w0, we are equipped to determine all future values of y.
We can write y1 as
y1=ϕy0+w1=ϕ(ϕy−1+w0)+w1=ϕ2y−1+ϕw0+w1.
Similarly y2 can be written as
y2=ϕ3y−1+ϕ2w0+ϕw1+w2,
and y100 can be written as
y100=ϕ101y−1+ϕ100w0+ϕ99w1+⋯+ϕw99+w100.
In general
yt=ϕt+1y−1+ϕtw0+ϕt−1w1+⋯+ϕwt−1+wt.
Another general form is
yt+j=ϕj+1yt−1+ϕjwt+ϕj−1wt+1+⋯+ϕwt+j−1+wt+j.
Effect of wt on yt+j is
∂wt∂yt+j=ϕj.
Points to note:
- The above equation is also called Impulse Response Function (IRF).
- When ∣ϕ∣<1, the system is stable. When ∣ϕ∣>1, the system is explosive.
- If ∣ϕ∣<1, the cumulative effect of one unit increase in wt on yt+j, where j∈{0,1,2,⋯}, is
j=0∑∞∂wt∂yt+j=j=0∑∞ϕj=1−ϕ1.
Example
Let the stream of future values be {yt,yt+1,yt+2,⋯} and the interest rate be r. Present value (PV) of the stream at time t is given by
PV=yt+1+ryt+1+(1+r)2yt+2+(1+r)3yt+3+⋯
Let β be the discount factor, hence
β=1+r1.
Therefore
PV=yt+βyt+1+β2yt+2+β3yt+3+⋯=j=0∑∞βjyt+j
Effect of one time increase
What would be the impact of increasing wt by one unit on PV while keeping wt+1,wt+2,…,wt+j unchanged?
∂wt∂PV=j=0∑∞βj∂wt∂yt+j=j=0∑∞βjϕj=1−βϕ1
provided ∣βϕ∣<1.
Effect of permanent change on PV
What would be the impact of a permanent change in w on PV?
A permanent change in w means that wt,wt+1,⋯, and wt+j would all increase by one unit.
Total change in PV can be calculated as follows
Total Change in PV=∂wt∂PV+j=1∑∞βj∂wt+1∂yt+j+j=2∑∞βj∂wt+2∂yt+j+⋯+j=j∑∞βj∂wt+j∂yt+j=j=0∑∞βjϕj+j=1∑∞βjϕj−1+j=2∑∞βjϕj−2+⋯+j=j∑∞βjϕ0=j=0∑∞βjϕj+βj=1∑∞βj−1ϕj−1+β2j=2∑∞βj−2ϕj−2+⋯+j=j∑∞βjϕ0
Effect of permanent change on yt+j
The effect on yt+j of a permanent change in w beginning in period t is given by
∂wt∂yt+j+∂wt+1∂yt+j+∂wt+2∂yt+j+⋯+∂wt+j∂yt+j=ϕj+ϕj−1+ϕj−2+⋯+ϕ+1.
When ∣ϕ∣<1, then
j→∞lim[∂wt∂yt+j+∂wt+1∂yt+j+∂wt+2∂yt+j+⋯+∂wt+j∂yt+j]=1+ϕ+ϕ2+⋯=1−ϕ1.