0
$\begingroup$

Given a price vector $(p_1,p_2,...,p_n)$ for some stock, then the corresponding return at $k$th day is described by $$ R_k = \frac{p_{k+1} - p_k}{p_k} $$

On the other hand, let $W_k$ be wealth at day $k$ induced by some trading algorithm. Then I have another ''return" (under some trading algorithm) given by $$ r_k = \frac{W_{k+1} - W_k}{W_k} $$

Assuming no risk-free rate. I want to use annualized Sharpe ratio to characterize my trading performance, then I got huge confusion for the following two possibilities: [The Sharpe Ratio formula below are fixed. ]

$$SR_1 = \sqrt{252} \frac{E[R_k] }{Std(R_k)}$$

$$SR_2 = \sqrt{252} \frac{E[r_k] }{Std(r_k)}$$

which one is the correct annualized Sharpe ratio? Any suggestion is appreciated.

$\endgroup$
3
$\begingroup$

First, you do not divide by the variance, but the standard deviation when calculating Sharpe ratios. Secondly, none of them are wrong, but $SR_1$ is the expected Sharpe ratio of the asset you are trading, and the second is the expected Sharpe ratio of your strategy. As a trader you care about the latter, but the first can be interesting to see if you actually beat just buying and holding the instrument in question.

What this means is that, notationally, you don't input the ex ante (expected) return in your Sharpe ratio calculation if you want to evaluate your results ex post (after they happened.) So you calculate your return series $r = \{r_1, r_2, r_3, \ldots, r_t\}$ and calculate the sample mean, $\hat{\mu}_{r}$ and similarly with the standard deviation $\hat{\sigma}_r$. This gives you the realized or ex post Sharpe ratio.

To make it explicit per Alex C's suggestion, calculating the ex post Sharpe ratios for both the strategy and the asset can be used to do what I describe in the first paragraph.

$\endgroup$
  • $\begingroup$ A comparison of the two ex-post Sharpe ratios will help answer the question: is the new trading strategy worthwhile or should you just have held the underlying asset passively, with less work and trouble on your part. $\endgroup$ – Alex C Feb 12 '17 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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