How to reproduce time series from simple moving average

Question

For a time series $X = \{X_1, X_2, ...\}$,

knowing its n-period simple moving average $SMA = \{SMA_n, SMA_{n+1}, ...\}$,

where $SMA_m = {X_m + X_{m-1} + ... + X_{m-(n-1)}\over n}$.

How to reproduce time series $X$ from $SMA$, if n is 30?

Answer 1

I'll assume that $X_m = 0\;\forall m<1$.

Let $S_m = nSMA_{m}$.

Consider the sequence $\left\{S_m \right\}$ for $n=3$. Then

$\begin{array}{*{20}llllll|ll} {{S_1} = }&{{X_1}}&{}&{}&{}&{}&{}&{{X_1} = {S_1}}&{} \\ {{S_2} = }&{{X_1} + }&{{X_2}}&{}&{}&{}&{}&{{X_2} = {S_2} - {S_1}}&{} \\ {{S_3} = }&{{X_1} + }&{{X_2} + }&{{X_3}}&{}&{}&{}&{{X_3} = {S_3} - {S_2}}&{} \\ {{S_4} = }&{}&{{X_2} + }&{{X_3} + }&{{X_4}}&{}&{}&{{X_4} = \left( {{S_4} + X_1} \right) - {S_3}}&{} \\ {{S_5} = }&{}&{}&{{X_3} + }&{{X_4} + }&{{X_5}}&{}&{{X_5} = \left( {{S_5} + X_2} \right) - {S_4}}&{} \\ \vdots &{}&{}&{}&{}&{}&{}& \vdots &{} \end{array}$

where it becomes evident that $$X_1 = S_1$$ $$X_m = S_m - S_{m-1}\;\;{\rm for }\;\; 2 \leq m \leq n$$ $$X_m = \left( S_m + X_{m-n}\right) - S_{m-1} \;\;{\rm for }\;\;n < m$$ You can also get an answer purely in terms of the $S$'s by solving the matrix equation (e.g. for $n=3$)

$$\left( {\begin{array}{*{20}{l}} 1&{}&{}&{}&{}&{} \\ 1&1&{}&{}&{}&{} \\ 1&1&1&{}&{}&{} \\ {}&1&1&1&{}&{} \\ {}&{}&1&1&1&{} \\ {}&{}&{}&{}& \ddots & \ddots \end{array}} \right)\left[ {\begin{array}{*{20}{c}} {{X_1}} \\ {{X_2}} \\ {{X_3}} \\ {{X_4}} \\ {{X_5}} \\ \vdots \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{S_1}} \\ {{S_2}} \\ {{S_3}} \\ {{S_4}} \\ {{S_5}} \\ \vdots \end{array}} \right]$$ where the matrix component is guaranteed to be full rank and thus invertible.