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Just a note

If E[] = Θ + b/n
and E[-j] = Θ + b/[n-1]
    then nE[] = nΘ + b
      and [n-1]E[-j] = [n-1]Θ + b

So, given that E[j] = nE[] - [n-1]E[-j]

    then E[j] = [nΘ + b] - [(n-1)Θ + b]
      = nΘ - [n-1]Θ = Θ