Weight vector of an exponential moving average?

Question

With weight vector I mean the vector with weights that you have to multiply the observations in the window that slides over your data with so if you add those products together it returns the value of the EMA on the right side of the window.

For a linear weighted moving average the formula for finding the weight vector is: (1:n)/sum(1:n) (in R code). This series of length n adds up to 1. For n=10 it will be

0.01818182
0.03636364
0.05454545
0.07272727
0.09090909
0.10909091
0.12727273
0.14545455
0.16363636
0.18181818

the numbers 1 to 10 / 55, with 55 the sum of the numbers 1 to 10.

How do you calculate the weight vector for an exponential moving average (EMA) of length n?

Wikipedia tells me:

if n is the length of the window, then alpha<-2/(n+1) and i<-1:n so EmaWeightVector<-((alpha*(1-alpha)^(1-i)))

Is this correct?

Even though the EMA is not really confined to a window with a start and an end, shouldn't the weights add up to 1 just like with the LWMA?

Answer 1

I believe the Wikipedia reference may refer to the Hunter (1986) but Wikipedia articles do morph over time. One aspect of the accepted answer that is less than satisfying is the code to divide by the sum.

The current article makes mention of this chart.

From https://en.wikipedia.org/wiki/Moving_average#/media/File:Exponential_moving_average_weights_N=15.png gives the code associated with it.

$N = 15;
$f = 1 - 2/($N+1);

$total = 0;
foreach $i (0..20) {
  $value = (1 - $f) * ($f ** $i);
  print "$i $value\n";
  $total += $value;
}
print "# total $total (should approach 1)\n";

Which in R is essentially:

N = 15;
a = 1 - 2/(N+1); #it is more customary to use a for the alpha rather than f
i=0:20 #increase this to a larger number to see how the sum approaches 1.0
value = (1 - a) * (a ^ i);
sum(value) #the sum eventually approaches 1.0 
plot(value);grid()

The above weights do eventually sum to 1 as they should without being forced...

I hope this may be of use.

Answer 2

ok the solution was: (in R)

emaweights<-function(m)
{
  alpha<-2/(m+1)
  i<-1:m
  sm<-sum((alpha*(1-alpha)^(1-i)))
  return(((alpha*(1-alpha)^(1-i)))/sm)  
}

Here the weights do add up to 1.

Answer 3

This will get you close to a closed form solution...

The weight at each vector is: $$\frac{(1-L)L^t}{1-L^T}$$

Where:

• $L$ is your smoothing constant; $(1-L)$ is often referred to as alpha
• $t$ is your time vector
• $T$ your last $t+1$

This doesn't exist anywhere else on the internet, so I had to invent it. It may therefore result in slightly different weights, but it works.