Good, but ...

• Generally:

  • E{log2(x)} ¹ log2(E{x})

• Our case:

  • E{log2mj} ¹ log2(E{mj})=aj +c

hypothesis

estimated

• Solution: introducing a correction factor g(j)
  • new random variable: yj = log2mj - g(j)
  • perform linear regression of yj on j

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