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