To recap:

1. We compare our portfolio to a "benchmark" portfolio, over a long time period.
2. We look at the ratio of the value of our $1.00 portfolio to the benchmark : call it R(n), after n years.
3. Adopting a utility function, U[x] = - x^-g, to measure our pleasure at having achieved a value x, we calculate U[R(n)] = - R^-g
4. We look at the expected value of our utility : E[U] = E[- R^-g].
5. We investigate this when n is large:
       Since R(n) is {(1+p)/(1+b)}ⁿ where we let p and b be the two annualized returns (after n years)
       then - R^-g = - {(1+p)/(1+b)}^-ng
       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^-g])
       Note that log(R^-g) = log({(1+p)/(1+b)}^-ng) = - n g 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^-g])

       Note that (again using our annualized returns, p and b) this would look like : g 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 g ... and n, of course, but we'll be letting n become infinite.
9. In particular, we note that : - e^{-n D} = - E[- R^-g]
       .. 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 g so as to maximize the limiting value of D(n) for n infinity.

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

Anyway, replacing -E[-R^-g] with E[R^-g] 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]] ≈ g log((1+p)/(1+b)) ≈ g (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 g 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 g (p - q) return by using [1A]:

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

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

[1C]       g E[p - q] - (1/2) g² StandardDeviation²[ p]
... since StandardDeviation[cx+d] = c² StandardDeviation[x] if c and d are constants

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

[1D]       g = 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.