Consider a discrete time series \(\{X_t | t = 0, 1, 2, \ldots\}\) where the \(X_t\) could be either deterministic (like samples from a sine wave) or a random variable. We can form a simple moving average from this time series by defining a new time series \(Y_t\), where:

\begin{equation*} Y_t = \frac12 \left(X_t + X_{t-1} \right) \end{equation*}

i.e. \(Y_t\) represents the average of the current value of \(X_t\) and the previous value, \(X_{t-1}\).

Using a moving average instead of the base series has many applications in the real world – signal processing, finance, etc., because it is usually considered an easy way to smooth the underlying signal \(X_t\).

This post shows how to invert this smoothing process, so that we can recover the underlying signal \(X_t\) from the smoothed signal \(Y_t\).

Inverting the simple moving average

To invert a simple moving average, first we define the shift operator \(B\), in terms of its action on a discrete time series:

\begin{equation*} B X_t = X_{t-1} \end{equation*}

that is, it shifts the time index of \(X\) back one step (\(t \to t-1\)). Therefore we can rewrite the moving average \(Y_t\) as:

\begin{equation*} Y_t = \frac12 (1+B)X_t \end{equation*}

In this form, we might guess that solving for \(X_t\) for would just be a matter of dividing both sides by \(\frac12 (1+B)\):

\begin{equation*} \frac{2}{1+B} Y_t = X_t \end{equation*}

However, since \(B\) is an operator and not a plain number, we need to make sense of what \(B\) being in the denominator means. The typical way to do this is to use the geometric series representation¹ to write the fraction as a infinite sum:

\begin{equation*} \frac{1}{1+B} = \sum_{k=0}^{\infty} (-1)^k B^k \end{equation*}

where \(B^k\) is just the application of \(B\) to the series \(k\) times:

\begin{equation*} B^k X_t = X_{t-k} \end{equation*}

Therefore, the solution should be understood as:

\begin{align*} X_t &= 2\sum_{k=0}^\infty (-1)^k B^k Y_t \\\
&= 2\sum_{k=0}^\infty (-1)^k Y_{t-k} \end{align*}

Generalization to higher-order moving averages:

If we write the original moving average as:

\begin{equation*} Y_t = \frac12 \sum_{k=0}^{1} X_{t-k} \end{equation*}

then we can generalize to a moving average of order \(q\), which is abbreviated MA(q):

\begin{equation*} Y_t = \frac{1}{q} \sum_{k=0}^{q-1} X_{t-k} \end{equation*}

Inverting this series is similar to the simple moving average. First, rewrite \(Y_t\) in terms of the shift operator \(B\):

\begin{equation*} \frac{1}{q}\sum_{k=0}^{q-1} X_{t-k} = \frac{1}{q} \left(\sum_{k=0}^{q-1} B^k\right) X_t \end{equation*}

Using our trick from earlier (but going in the opposite direction), the term in parenthesis can be written as a fraction:

\begin{equation*} \sum_{k=0}^{q-1}B^k = \frac{1-B^{q}}{1-B} \end{equation*}

which allows us to solve for \(X_t\) by multiplying both sides by the reciprocal:

\begin{equation*} X_{t} = q\frac{1-B}{1-B^q}Y_t \end{equation*}

and from here, we expand the denominator as an infinite sum:

\begin{align*} X_t &= q (1-B)\sum_{k=0}^{\infty}B^{kq} Y_t \\\
&= q \sum_{k=0}^\infty (B^{qk} - B^{qk + 1}) Y_t \\\
&= q \sum_{k=0}^\infty (Y_{t-qk} - Y_{t-(qk+1)}) \\\
\end{align*}

Sanity check

Sanity checks are always helpful. Believe it or not, I actually made a mistake the first time I wrote this. So, does this reduce to the previous solution in the case \(q = 2\)? The answer is yes:

\begin{align*} X_t &= 2\sum_{k=0}^\infty (Y_{t-2k} - Y_{t-2k-1}) \\\
&= 2\left(Y_{t} - Y_{t-1} + Y_{t-2} - Y_{t-3} + \ldots \right) \\\
&= 2\sum_{k=0}^\infty (-1)^k Y_{t-k} \end{align*}

QED