I am trying to find out more about the convergence of moving averages. Unfortunately, the use of moving averages in stock trading has polluted my searches rendering them fruitless. After looking through old posts, hopefully this SE is the best place to ask my question:
Suppose we have an input of three real numbers $t_1$, $t_2$ and $t_3$ such that 0 $\leq$ $t_1$, $t_2$, $t_3$ < 1 and an output moving average $t_n$ = $ \frac{1}{3} (t_{n-1}+t_{n-2}+t_{n-3})$. For example:
[0.2 0.4 0.3]
0.2 [0.4 0.3 0.3]
0.2 0.4 [0.3 0.3 0.3333...] and so on
Is there a proof that, for any input $t_1$, $t_2$, $t_3$, over time all numbers in $t_n$ converge to a number between 0 and 1? If so is there a formula for which value it converges to?
This is a typical example of linear recurrence, so we may adopt the standard approach. The characteristic equation of the recurrence relation is given by
$$ x^3 = \tfrac{1}{3}(x^2 + x + 1). $$
This equation has 3 zeros $x = 1, \alpha, \bar{\alpha}$, where $\alpha = \frac{-1+i\sqrt{2}}{3}$. Then the general solution of the equation takes the form
$$ t_n = A + B \alpha^n + C \bar{\alpha}^n. \tag{*} $$
Indeed you may easily check, using the characteristic equation, that $\text{(*)}$ indeed solves the given recurrence relation. Conversely, any solution is determined by knowing 3 leading terms $t_1, t_2, t_3$, and we can always find $A, B, C$ for which $\text{(*)}$ produces those prescribed leading terms. More precisely, $A, B, C$ are given by the unique solution of the following system of equations
$$ \begin{pmatrix} t_1 \\ t_2 \\ t_3 \end{pmatrix} = \begin{pmatrix} 1 & \alpha & \bar{\alpha} \\ 1 & \alpha^2 & \bar{\alpha}^2 \\ 1 & \alpha^3 & \bar{\alpha}^3 \end{pmatrix} \begin{pmatrix} A \\ B \\ C \end{pmatrix} $$
Now, since $|\alpha| = |\bar{\alpha}| = \frac{1}{\sqrt{3}} < 1$, we know that $\alpha^n \to 0$ and $\bar{\alpha}^n \to 0$. So it follows that the limit of $\text{(*)}$ is simply $A$. Then, solving the above equation, we get
$$ \lim_{n\to\infty} t_n = A = \frac{t_1 + 2t_2 + 3t_3}{6}. $$
Moreover, the difference between $t_n$ and the limit $A$ decays at least exponentially fast.