Exponential Moving Average

When we want to filter noise and have an idea about the trend of a time series, we usually apply linear or exponential moving average (MA) depends on the noise is time independent or not. the latter assumes current value is affected by last one, thus we can define the relation as

$$ v_t = \beta p_t + (1 - \beta)v_{t-1} $$

Where $p_t$ and $v_t$ are the pratical/weighted value at time $t$ respectively, $\beta$ is the weighting for current pratical value and thus $(1-\beta)$ is for the last weighted value.

Recurrence Relation¶

Let's expand $v_{20}$

$$ \begin{align*} v_{20} &= \beta p_{20} + (1-\beta) v_{19} \\ &= \beta p_{20} + (1-\beta) [\beta p_{19} + (1-\beta) v_{18}] \\ &= \beta p_{20} + \beta (1-\beta) p_{19} + (1-\beta)^2 [\beta p_{18} + (1-\beta) v_{17}] \\ &= \cdots \\ &= \beta [ p_{20} + (1-\beta) p_{19} + (1-\beta)^2 p_{18} + \cdots + (1-\beta)^k p_{20-k} ] \end{align*} $$

thus we get the recurrence relation $$ v_t = \beta \vec{P} \cdot \vec{W} $$ where $$ \begin{align*} \vec{P} &= [p_t &&,p_{t-1} &&,p_{t-2} &&\cdots &&,p_1 &&,p_0] \\ \vec{W} &= [\beta &&,\beta (1-\beta) &&,\beta (1-\beta)^2 &&\cdots &&,\beta (1-\beta)^{t-1} &&,\beta (1-\beta)^t] \end{align*} $$

In [1]:

import matplotlib.pyplot as plt

beta = 0.3
steps = 20
T = range(0, 21)
W = [beta * (1-beta)**(steps-t) for t in T]
plt.plot(T, W, 'o--')
plt.annotate(r'$\beta$', xy=(T[-1]-0.5, W[-1]), xytext=(T[-1]-3, W[-1]), arrowprops={'arrowstyle': '->'})
plt.annotate(r'$\beta(1-\beta)$', xy=(T[-2]-0.5, W[-2]), xytext=(T[-2]-5, W[-2]), arrowprops={'arrowstyle': '->'})
plt.annotate(r'$\beta(1-\beta)^2$', xy=(T[-3]-0.5, W[-3]), xytext=(T[-3]-5, W[-3]), arrowprops={'arrowstyle': '->'})
plt.title("Weighting on $p_t$")
plt.xlabel("T")
plt.ylabel("W")
plt.xticks(range(0,21,2))
plt.grid(linestyle='--')
plt.show()

In this $\beta = 0.3$ sample, $v_{20}$ has a contribution from $p_{20}$ with factor 0.3 ($\beta$), and drop exponentially, 0.21 on $p_{19}$ and 0.05 on $p_{15}$ already.

Discusstion on $\beta$¶

Using Compound interest as an example, known that

$$ \lim_{n \to \infty} (1 + \frac{r}{n})^n = e^r $$

To get our previous formula, let $r = -1$ and $\beta = 1 / n$, then

$$ \lim_{\beta \to 0} (1 - \beta)^{\frac{1}{\beta}} = \frac{1}{e} \approx 0.3678 $$

Which means the weighting drop to about $1/3$ after $1/\beta$ periods.

Unlike fixed window size in linear MA, we can easily define 1 week or 1 month MA. Exponential MA affacets by all previous elements. By convention we use $1/e$ as a reference and says it's a $1/\beta$ period EMA. e.g. To calculate 1 week EMA of stock price, we let $1/\beta = 5$ thus set $\beta$ to 0.2.