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

Question

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

Answer 1

I think some some terminology got mixed up here.

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 $S = \text{sgn}(\lambda_p) |\lambda_p|=\lambda_p$.