Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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

share|improve this question
1  
Please provide a link to the complete definition of the Stutzer Index for the sake of completeness –  SRKX Sep 23 '13 at 15:00
    
I found a better definition here: activetradermag.com/index.php/c/Article_Follow-ups/d/… –  John Sep 26 '13 at 17:25
add comment

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

share|improve this answer
    
very clear explanation...any help on below: If I have only 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) –  purnendumaity Sep 29 '13 at 22:20
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.