Linear First Order Difference Equations (Iterative/General Method)

[rabbitchannel]

Homework Statement

I am almost done with a chapter all about this topic and this type of question is the only one I can't get. This is linear first order difference equations. The question is:

Given the unemployment U_t equation:

U_t = $$\alpha$$ + $$\beta$$ U_t-1
$$\alpha$$, $$\beta$$ > 0

b. Suppose that there are occasional shocks to the demand for labor causing shifts in U_t. The modified equation for U_t becomes:

U_t = $$\alpha$$ + $$\beta$$ U_t-1 + e_t

where e_t varies over time. Show that the solution to the modified equation is:

U_t = $$\beta$$^tU₀ + [tex]\frac{\alpha(1-\beta^t)}{1-\beta}[/tex] + e₁$$\beta$$^t-1 + e₂$$\beta$$^t-2 + ... + e_t-1$$\beta$$ + e_t

Don't know how to fix that there. It should be (1-$$\beta$$^t)

Homework Equations

General Method:
P_c + P_p = General method

Y_t = (Y₀ - $$\frac{c}{1+a}$$)(-a)^t + $$\frac{c}{1+a}$$

I've also got the derived formula for supply and demand but that requires two functions.

The Attempt at a Solution

Ok, I can't get the iteration. This is what I've tried:

U_t = $$\alpha$$ + $$\beta$$U_t-1 + e_t

U_t+1 = $$\alpha$$ + $$\beta$$U_t + e_t+1

After this point I don't know what to do. I tried to do this:

U_t+1 = $$\beta$$($$\alpha$$ + $$\beta$$U_t + e_t+1) + $$\alpha$$ + e_t+1

Basically multiplying the whole equation by $$\beta$$ then adding: $$\alpha$$ + e_t+1. Once I do it for 3 periods I can determine the general function but it is different from the given one. I lack the 1-$$\beta$$ on that denominator. I can solve other equations but have trouble when something else, such as e_t+1 is added. I've also used the general method but it also turns out different. I was under the impression that I can use the iterative and general solutions for any first order linear difference equation. Am I wrong?

Any help would be greatly appreciated. Thanks!

P.S. I would like to thank the system for auto-logging me out while trying to preview my first ever post, thereby deleting a chunk of what I wrote. Good thing I saved. :\

 

[Nino MMP]

The solution to this problem is actually relatively simple. Start with the given equation:Ut = \alpha + \beta Ut-1 + etThen, subtract Ut-1 from both sides to get:Ut - Ut-1 = \alpha + \beta Ut-1 - \beta Ut-1 + etSimplifying, we have:Ut - Ut-1 = \alpha + etNow, let's rearrange and solve for Ut:Ut = Ut-1 + \alpha + etNow, let's use the same technique to get the next period's equation:Ut+1 = Ut + \alpha + et+1We can continue this pattern of rearranging and substituting to get the general equation:Ut = \betatU0 + \frac{\alpha(1-\betat)}{1-\beta} + e1\betat-1 + e2\betat-2 + ... + et-1\beta + etThis is the solution to the modified equation.