Analytic form of power series model?

Question

The Setup

I am dealing with the recursive equation $$w_t = \alpha + (1-\alpha)w_{t-1}, $$ with $0 < \alpha < 1$ and some starting point $w_0$. I am trying to prove an expression for $w_t$ that does not rely on $w_{t-1}$.

So for instance, $$w_1 = \alpha + (1-\alpha)w_0,$$ and \begin{align*} w_2 &= \alpha + (1-\alpha)[\alpha + (1-\alpha)w_0] \\ &= \alpha(1 + (1-\alpha)) + (1-\alpha)^2w_0. \end{align*} At this point I know that $$\alpha(1 + (1-\alpha)) + (1-\alpha)^2 = 1$$ so I can rewrite $w_2$ as $$w_2 = 1-(1-\alpha)^2 + (1-\alpha)^2w_0.$$

The Problem

I am trying to generalize the result to $w_t$, but am having trouble proving that $$\alpha(1 + (1-\alpha)^1 + \ldots + (1-\alpha)^{t-1}) + (1-\alpha)^t = 1$$ for any positive integer $t$.

Any strategies on how to prove this would be much appreciated!

Answer 1

Even if simple, I shall go through the steps. We have $$w_t = \alpha + (1-\alpha)w_{t-1}$$ and what is not pleasant is the constant term. So, let $w_t=u_t+k$ and replace $$u_t+k=\alpha + (1-\alpha)u_{t-1}+(1-\alpha)k$$ To get rid of the constant, write $$k=\alpha+(1-\alpha)k\implies k=1$$ and then $$u_t=(1-\alpha)u_{t-1}\implies u_t=C (1-\alpha)^{t-1}\implies w_t=1+C (1-\alpha)^{t-1}$$ But, for $t=0$, you have $$w_0=1+C \implies C=w_0-1$$

Answer 2

First try to solve $a_t = (1 - \alpha) a_{t - 1}$ with the guess $a_t = Cr^t$. Then try $w_t = D + a_t$ and solve for $D$. This is the usual approach to solving linear recurrence relations. Let me know if you need more guidance.

Answer 3

If the sequence converges, it converges to the solution of

$$w_\infty=\alpha+(1-\alpha)w_\infty$$ which is $$w_\infty=1.$$

Then by subtraction, the recurrence can be rewritten as

$$w_t-w_\infty=(1-\alpha)(w_{t-1}-w_\infty)$$

which, by induction has the obvious solution

$$w_t-w_\infty=(1-\alpha)^t(w_0-w_\infty)$$

or

$$w_t=(1-\alpha)^t(w_0-w_\infty)+w_\infty=(1-\alpha)^t(w_0-1)+1.$$

---

Technical note: the sequence does not need to converge for the computation to be valid, it works for any $\alpha$. (If it does not converge, $w_\infty$ is an unstable equilibrium point.)