Gradual mean shift intervention

Question

In Cryers textbook: Time Series Analysis: With Applications in R, I came across this:

$$ m_t = \delta m_{t-1} + wS^d_{t-1}\\ \text{where,} \, \,S_t^d = \begin{array}{cc} \{ & \begin{array}{cc} 1 & t \geq d \\ 0 & \text{else} \end{array} \end{array} $$

Cryer explains, "After some algebra, it cam be shown that : "

$$ m_t = \begin{array}{cc} \{ & \begin{array}{cc} w\frac{1-\delta^{t-d}}{1-\delta} & \text{for } t > d \\ 0 & \text{otherwise} \end{array} \end{array} $$

where $|\delta| \leq 1$ represents the half life $\frac{log(0.5)}{log(\delta)}$.

I'm not seeing how he got this. If I distribute, and insert an $S_t^d$: $$ m_t = \delta m_{t-1} + wS_{t-1}^d \rightarrow\\ m_t (1-\delta B) = w S_t^d\rightarrow\\ m_t = \frac{w}{(1-\delta B)}S_t^d\rightarrow\\ m_t = w (1 + \delta B + \delta^2 B^2 + \delta^3 B^3 + ... ) S_t^d $$

Thus,

$$ m_{d-1} = 0\\ m_d = w * (1)\\ m_{d+1} = w * (1 + \delta) \\ m_{d+2} = w * (1 + \delta + \delta^2)\\ m_{d+3} = w * (1 + \delta + \delta^2 + \delta^3) $$

Cryer explains that this model is supposed to represent a gradual shift to a new mean. What is the new mean? Is it $\, lim_{\, m \rightarrow \infty} w * (1 + \delta + \delta^2 + ... + \delta^m)$?

Also, what is the "after some algebra" Cryer uses to transform the equation?

Answer 1

If it is assumed or known that $m_s=0$ for some $s\lt d-1,$ then $S_s^d=0$ and the recurrence relation gives

$$m_{s+1} = \delta m_s + w S_s^d= 0.$$

If $0 \lt d,$ then in the same manner we may continue the process to deduce

$$0 = m_s = m_{s+1} = \cdots = m_{d} = 0.$$

When $t=d+1,$ though, for the first time $S_d^d = 1$ and we obtain

$$m_{d+1} = \delta m_d + w S_{d}^d = \delta(0) + w = w.$$

Next, because $S^d_{d+1}=1,$ the recurrence gives

$$m_{d+2} = \delta m_{d+1} + w S_{d+1}^d = \delta w + w = w(1 + \delta).$$

This is a recurrence of the form "the next value is a constant times the previous value plus another constant." Its general solution is derived at https://stats.stackexchange.com/a/444788/919. That presents the kind of algebra someone would need to do if they did not know the answer in advance.

However, if you only care about checking a putative solution, such as the one given in the book, you can establish it inductively by supposing the solution holds at some time $t \gt d$ and verifying the solution holds for the next time $t+1.$ This is done, of course, by using the recurrence relation to relate $m_{t+1}$ to $m_t:$

$$\eqalign{m_{t+1} &= \delta m_t + w = \delta \left(w \frac{1 - \delta^{t-d}}{1-\delta}\right) + w\\ &= w\left[\delta \left(\frac{1 - \delta^{t-d}}{1-\delta}\right) + 1 \right] \\ &= w \frac{\delta - \delta^{t-d+1} + 1 - \delta}{ 1 - \delta} \\ &= w \frac{1 - \delta^{(t+1)-d}}{1-\delta} , }$$

as claimed. The first line plugged the putative solution into the recurrence, and then everything after the first line was the "algebra" referred to: first $w$ was factored out (to make the answer look like a multiple of $w$) and then the fractions were added and simplified.

Finally, when $|\delta| \lt 1,$ the numerator of the fraction approaches $1$ (exponentially quickly), giving

$$\lim_{t\to\infty} m_t = w \frac{\lim_{t\to\infty} \left(1- \delta^{t-d}\right)}{1-\delta} = w \frac{1}{1-\delta} = \frac{w}{1-\delta}.$$

If you consider a (decaying) exponential change to be "gradual," then everything is as claimed in the text. When $|\delta| \approx 1,$ that is a reasonable characterization of the situation.