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Expression for mean and variance in a running stream

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  • Last Updated : 22 Feb, 2021
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Let we have a running stream of numbers as x1,x2,x3,…,xn.

The formula for calculating mean and variance at any given point is given as : 

  • Mean = E(x) = u = 1/ni=1n xi
  • Standard Deviation = s = 1/ni=1n (xi – u) 2
  • Variance = s2

However, it would be a very slow approach if we calculate these expressions by looping through all numbers each time a new number comes in.

Effective solution

s2 = 1/ni=1n (xi - u) 2

    = 1/n (∑i=1n xi2 + ∑i=1n u2 - 2u ∑i=1n xi)

    = 1/n (∑xi2 + nu2 - 2u ∑xi)

    = ∑xi2/n + u2 - 2u ∑xi/n

    = ∑xi2/n - u2

    = E(x2) - u2

    = E(x2) - [E(x)]2

Therefore, in this implementation, we have to maintain a variable sum of all the current numbers for mean and maintain variable sum2 of all the current numbers for E(x2) and we have to maintain another variable n for the count of numbers present.

Python code for the implementation :

Python3




sum=0    # To store sum of stream
sumsq=0  # To store sum of square of stream
n=0      # To store count of numbers
while(True):
      
    x=int(input("Enter a number : "))
      
    n+=1
    sum+=x
    sumsq+=(x*x)
      
    #Mean
    mean = sum/n
    #Variance
    var = (sumsq/n) - (mean*mean)
    print("Mean : ",mean)
    print("Variance : ",var)
    print()

Input and corresponding output : 

Enter a number : 1
Mean :  1.0
Variance :  0.0

Enter a number : 2
Mean :  1.5
Variance :  0.25

Enter a number : 5
Mean :  2.6666666666666665
Variance :  2.8888888888888893

Enter a number : 4
Mean :  3.0
Variance :  2.5

Enter a number : 3
Mean :  3.0
Variance :  2.0

Thus, we can compute mean and variance of a running stream at any given point in constant time.

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