# What is the correct Stutzer index and Sharpe ratio relation, assuming a normal returns distribution?

Assuming the returns distribution is normal, then there is a relation between Stutzer index and Sharpe ratio.

However, I found in the following paper 2 different equation:

• Paper I (page 10-11)‎ where it is mentioned Stutzer index (Ip) is half of square of the Sharpe ratio.

• Paper II (page 8) where it is mentioned Stutzer index is equal to the Sharpe Ratio.

Can somebody tell, which one is correct?

Also if I have ony 12 monthly return series is it meaningful to calculate Stutzer index? (most of the implemented algorithms I'v seen so far are on daily returns of at least 100-120 observations)

Stutzer index definition: http://www.investopedia.com/terms/s/stutzerindex.asp

Michael J. Stutzer original paperlink: http://papers.ssrn.com/sol3/papers.cfm?abstract_id=239540

Let $r_t$, $t=1,\ldots,T$ be a series of iid excess returns with the estimated mean excess return $\bar{r}= \sum_{t=1}^Tr_t$. Then the Stutzer Index $S$ is defined as $S=\frac{|\bar{r}|}{\bar{r}}\sqrt{2I_p}$ with $I_p$ being the "Stutzer Information Statistic", $I_p=\max_\theta -\log(\frac{1}{T}\sum_{t=1}^T \text{e}^{\theta r_t})$. In the normal case John's reference tells us that $I_p = \frac{1}{2}\lambda_p^2$ where $\lambda_p$ is the Sharpe Ratio.
In this case, the Stutzer Information Statistic $I_p$ is obviously half of the squared sharpe ratio.
The Stutzer Index $S$ on the other hand is equal to the sharpe ratio:
Since $\frac{|\bar{r}|}{\bar{r}} = \text{sgn}(\lambda_p)$ and $\sqrt{2I_p} = |\lambda_p|$ it follows that $S = \text{sgn}(\lambda_p) |\lambda_p|=\lambda_p$.
• I doubt that it makes a lot of sense with only twelve data points but opinions can vary. Some people may argue that its better to calculate something than nothing. If you calculate it, be sure to compare it with the sharpe ratio. One thing I did not mention above is that the equality with the sharpe ratio only holds when we consider the expectations (instead of approximating them by means as in $I_p$ above). So even for this comparison you would want to have more returns... – vanguard2k Sep 30 '13 at 6:46