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  1. We start with a function f(t) where t (which we can think of as time)
    lies in the interval [0,T]
  2. We subdivide this interval by equally spaced points t0=0, t1, t2, t3 ... tN
    where Dt = tk+1 - tk   (k = 0,1,2, ... N-1)
  3. We construct the so-called Riemann Sum:
          f(t1)Dt+ f(t2)Dt + f(t3)Dt + ... + f(tN)Dt = S f(tk)Dt

Figure 1

[1]       S f(tk)Dt f(t) dt   as Dt 0

  1. Suppose returns r have a Mean M[r] and Variance V[r].
  2. If the returns r are lognormally distributed, then r = ex where the x's are normally distributed.
  3. Suppose the Mean and Variance for the x's are M[x] and V[x].
  4. Because of the lognormal association, the relation between M[r], V[r], M[x] and V[x] is
          V[x] = log[1+V[r]/{1+M[r]}2]
          M[x] = log[1+M[r])]- V[x]/2
  5. Hence eM[x] = (1+M[r]) e- V[x]/2 = (1+M[r]) [1+V[r]/{1+M[r]}2]-1/2
  6. The Median return is then r = eM[x] - 1 = (1+M[r]) [1+V[r]/{1+M[r]}2]-1/2 - 1
  7. For small M[x] and V[x], r is approximately
    r = (1+M[r]) [1- (1/2)V[r]] - 1 which is approximately r = M[r] - V[r]/2.
EXP(m+s2/2) = M
EXP(2m+s2) [EXP(s2)-1] = S2
Solving, we get the Magic Formula:
m = LOG(M) - s2/2
where s2 = LOG(1 + S2/M2)