To recap:

1. We compare our $1.00 portfolio to a "benchmark" portfolio, over a long time period.
2. We look at the ratio of the value of our portfolio to the benchmark, namely P(n) / B(n) : call this ratio R(n), after n years.
3. Adopting a utility function, U[x] = - x^-γ, to measure our pleasure at having achieved a value x, we calculate U[R] = - R^-γ
4. We look at the expected value of our utility : E[U] = E[- R^-γ] ... where R means R(n).
5. We investigate this when n is large:
       If p and b are the two annualized returns (after n years), then R(n) is {(1+p)/(1+b)}ⁿ
       then - R^-γ = - {(1+p)/(1+b)}^-nγ
       This is negative and would tend to zero (as n increases) if p > b.
       Note that p > b means our portfolio does better than the benchmark  
6. To avoid this zero limit, we take a logarithm, like so : log(-E[- R^-γ])
       Note that log(R^-γ) = log({(1+p)/(1+b)}^-nγ) = - n γ log((1+p)/(1+b))
       We'd expect this to head to - infinity as n infinity ... if p > b.
7. We massage this to avoid the infinite limit and introduce a modified measure of this variable by dividing by (-n), like so :
       (-1/n) log[-E[U]] = (-1/n) log(-E[- R^-γ])

       Note that (again using our annualized returns, p and b) this would look like : γ log((1+p)/(1+b))
       so we might expect a finite, positive, non-zero value as n infinity.

8. We now introduce a variable D(n) defined like so: D = (-1/n) log(-E[U])
       We'd expect this variable D to depend upon the two annualized returns and γ ... and n, of course, but we'll be letting n become infinite.
9. In particular, we note that : - e^{-n D} = - E[- R^-γ]
       .. so D is a kind of measure of how rapidly we approach a limit ... a rate of decay to that limit.
10. Finally, we choose γ so as to maximize the limiting value of D(n) for n infinity.

>Why all the negative signs? Isn't -E[-R^-γ] the same as E[R^-γ] ?
Yes, but we wanted a standard type of increasing utility function, namely -x^-γ

Anyway, replacing -E[-R^-γ] with E[R^-γ] we have:

Stutzer Index =

>Mamma mia! Does anybuddy actually use that guy?
How would I know? I just think it's fun, don't you?
>Certainly not!

Stutzer and Sharpe

We saw, above, that the D-values depend upon the portfolio returns and these are random.
There's a neat formula which approximates the annualized returns of a portfolilo (see this) and it goes like so:

[1A]       (Annualized Return) ≈ (Average Return) - (1/2)(Standard Deviation)²

In #7, above, we obtained this:

      (-1/n) log[-E[U]] ≈ γ log((1+p)/(1+b)) ≈ γ (p - q)
... since log((1+p)/(1+b)) = log(1+p) - log(1+q) ≈ p - q   because log(1+ x) ≈ x for small x.

Now that p-guy is a random variable (and, unless you're talking about a constant, risk-free return, so is b).
So this repesents a series of excess returns, amplified by a user-selected factor γ which is a measure of your delight in achieving an excess return.

>Delight? You mean your personal utility assignment, right?
Yes. Anyway, we can incorporate volatility into this γ (p - q) return by using [1A]:

[1B]       Average [γ (p - q)] - (1/2)StandardDeviation²[γ (p - q)]

Now suppose that b is a constant, risk-free return (a la Sharpe Ratio). Then we'd get:

[1C]       γ E[p - q] - (1/2) γ² StandardDeviation²[ p]
... since StandardDeviation[cx+d] = c² StandardDeviation[x] if c and d are constants

This is a simple quadratic in γ and has a maximum value at:

[1D]       γ = E[p - q] / StandardDeviation²[ p]

If we substitute this maximizing value for gamma, we get:

[1E]       (-1/n) log[-E[U]] ≈ (1/2) { E[p - q] / StandardDeviation }² = (1/2) {Sharpe Ratio}²

>Mamma mia!
My sentiments exactly.

Stutzer math

Note that the Stutzer Index is:

which is:
    the maximum (over γ) of
    the limit (as n infinity) of
    (-1/n) times
    the logarithm of the Expected value (or average) of   [ (Portfolio Value)/(Benchmark Value) ] ^-γ

The last two step involves the logarithm of an average.

That's way too complicated for us, but it'd be very neat (mathematically speaking) if,
instead of taking the logarithm of an average, we took the average of the logarithms  

That is, instead of
[A]       log{(1/m) [z(1) + z(2) + ... + z(m)]} ... that's the log of the average or the log of the Arithmetic Mean
we consider
[B]       (1/m){ log(z(1)) + log(z(2)) + ... + log(z(m))} ... that's the average of the logs or the log of the Geometric Mean  

We'll play with this idea:

1. Let z(k) = 1 + r(k) where r(k) is small.
2. From [A], the log(average) is:
         log{(1/m) [z(1) + z(2) + ... + z(m)]} = log{(1/m)Σ (1+r(k))} = log{1 + (1/m)Σ r(k)} ≈ (1/m)Σ r(k) - (1/2){(1/m)Σ r(k) }²
         since log(1+x) ≈ x - (1/2)x² when x is small.
3. From [B], the average(log) is:
         (1/m){ log(z(1)) + log(z(2)) + ... + log(z(m))} = (1/m)Σ log(1 + r(k)) ≈ (1/m)Σ r(k) - (1/2)(1/m)Σ r(k)²
         since log(1+x) ≈ x - (1/2)x² when x is small.
4. Staring intently at the two expressions, we conclude that:
         log(average) - average(log) ≈ (1/2) Average[r(k)²] - (1/2) Average²[r(k)]
5. In fact, Average[r(k)²] - Average²[r(k)] = StandardDeviation²[r(k)] ... see this.
6. We now have:
         log(average) ≈ average(log) + VAR[r(k)]
         where we're talking about variables z = 1 + r and r is small and where VARiance = StandarDeviation².
7. For our variables, namely z(k) = R(k)^-γ, we note that:
         average(log) = average(log(R(k)^-γ)) = average(-γ log(R(k))