Explicit formula for EWMA(n) in Python

Question

I have the following pandas.core.series.Series: x=pd.Series([17.39, 8.70, 2.90, 1.45, 1.45]) for t=0,1,2,3,4. When I try x.ewm(span=2).mean() I get the following results:

17.39, 10.87, 5.35, 2.71, 1.86

My understanding is that .ewm.mean() uses the following explicit formula:

y[t] = (1 - alpha) * y[t-1] + alpha * x[t], where alpha = 2/(span+1)

y[0] = x[0]

Using the above formula:

EMWA[0] = x[0] = 17.39

EMWA[1] = (1-(2/(2+1)))*17.39 + (2/(2+1))*8.7 = 11.59 which is different to 10.87.

EMWA[2] = (1-(2/(2+1)))*10.87 + (2/(2+1))*2.9 = 5.55 which is different to 5.35.

EMWA[3] = (1-(2/(2+1)))*5.35 + (2/(2+1))*1.45 = 2.75 which is different to 2.71. etc..

Could you please help me understand where these differences coming from? What I am missing or doing wrong here? Thank you.

For those jumping from R to Python (like me), in the following link you will find the difference between TTR::EMA(x,n=2) which simply computes the first non-missing value as the arithmetic mean of the first n variables and then recursively use y[t] = (1 - alpha) * y[t-1] + alpha * x[t] (for t>n), and the way x.ewm(span=2, adjust = False).mean() uses the y[0]=x[0] and then recursively y[t] = (1 - alpha) * y[t-1] + alpha * x[t] (for t>=1). stackoverflow.com/questions/12557697/… — Telis
– Telis, Commented May 10, 2023 at 6:48

Ben Grossmann · Accepted Answer · 2023-05-09 17:58:20Z

The formula that you're using is correct for x.ewm(span=2, adjust = False).mean(). Here is some simple code that correctly replicates the behavior under the default adjust = True setting.

span = 2
alpha = 2/(span + 1)

num = 0
den = 0
for b in x:
    num = (1 - alpha)*num + b
    den = 1 + (1 - alpha)*den
    print(num/den)

For more, see the documentation for ewm and look at the description for the adjust parameter.

The documentation gives you a summation formula for the adjust = True version for the EW function. If you want a comparable formula for the adjust = False version, it turns out that EMWA[n] is given by the result of

result = (1-alpha)**n * x[0] + alpha * sum((1-alpha)**(n-k) for k in range(1,t+1))

That is,

                      n                      
                    ____                     
                    ╲                        
       n             ╲           (n - k)     
(1 - α)  ⋅ x  + α ⋅  ╱    (1 - α)        ⋅ x 
            0       ╱                       k
                    ‾‾‾‾                     
                    k = 1

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