sumx += new-old
sumx2 += new²-old²
stdev = sqrt((sumx2/5000)-(sumx/50
but be carfull that round off errors don't make this negative
Main Topics
Browse All TopicsDoes anyone know of an incremental approximation for standard deviation? I currently use single pass [Knuth] for calculation but am willing to trade some accuracy for better than O(n) performance
Ex:
I have 5000 values, I have a standard deviation for them. I add a new value or remove an old value ... what is the new deviation?
Cheers,
Greg
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by: gregoryyoungPosted on 2007-09-13 at 11:36:32ID: 19886003
btw: I am 99.99% sure (but i have not sufficiently proven) that for an exact value I need O( n ) ... to be clear I am looking for an incremental approximation