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I have only one return for calculating sharp ratio. As you know, we should calculate standard deviation of returns and standard deviation of one item is 0. Suppose that the single return is 0.1 and the risk less return is 0.2. How can I calculate sharp ratio for these two inputs?

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closed as off-topic by Richard, LocalVolatility, amdopt, Quantuple, msitt Apr 16 '17 at 18:55

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    $\begingroup$ Clearly you can't. One way would be to look at similar asset which is correlated with yours and use that asset's volatility to calculate SR. $\endgroup$ – rbm Apr 10 '17 at 9:47
  • $\begingroup$ @rbm. I'm designing a trading system winch optimizes different parameters to generate buy and sell signals. In some severe cases, there is only one buy and sell signal, so I have only one return. What do you think about this case? The Sharpe ratio is my cost function in this system so I can't ignore it in this special case. $\endgroup$ – user2991243 Apr 10 '17 at 9:49
  • $\begingroup$ Personally I'd not handle this case at all, as it looks like you're describing an asset in highly illiquid market. Obviously I do not know all the details but sounds like I wouldn't even produce a signal $\endgroup$ – rbm Apr 10 '17 at 11:45
  • $\begingroup$ if you have only one buy and sell signal for a period of say 100 days theb your strategy has many 0 returns and the returns caused by the signal. $\endgroup$ – Richard Apr 10 '17 at 11:50
  • $\begingroup$ @Richard. Can you elaborate this more? What is your idea for handing this problem? As mentioned, this is cost function of my algorithm. $\endgroup$ – user2991243 Apr 10 '17 at 11:54
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For a single period return, the squared value of that return approximates variance (i.e., the absolute value approximates the standard deviation).

Standard deviation is defined thus:

$$\sigma_X = \sqrt\frac{\Sigma_1^N\mathbb{E}[X-\mu_x]^2}{N}$$

For a non-drifting process, $\mu_x = 0$. Also, in our scenario, $X = (r_a - r_m)$ and $N = 1$.

Therefore, an approximation of the Sharpe ratio should be:

$$S = \frac{r_a-r_b}{\sqrt{\mathbb{E}[r_a-r_m]^2}} \approx \frac{r_a-r_m}{\mid r_a-r_b\mid} $$

Using $r_a = .1$ and $r_b = .2$, $S$ should equal $-1$.

If you need annualize the returns or standard deviation, just remember that logarithmic returns scale with $T$ and logarithmic standard deviation should scale with $\sqrt{T}$. If operating under the assumption that returns are percentages, one must convert these into logarithmic returns first and/or use geometric compounding rules.

For a related thread see: Why is $dS/S$ an estimate of realized volatility?

Addendum: A perhaps pertinent point that I missed in my original pass through is that the mean absolute error is related to the expected standard error by a factor of ${\sqrt {2/\pi }}=0.79788456\ldots$, i.e.,

For the normal distribution, the ratio of mean absolute deviation to expected standard error is: $w=\frac{ E|X| }{ \sqrt{E(X^2)} } = \sqrt{\frac{2}{\pi}}$.

Furthermore, the median absolute error is connected with standard error by the following:

${\displaystyle {\hat {\sigma }}=k\cdot \operatorname {MAD} ,\,}$ where k is a constant scale factor, which depends on the distribution.

For normally distributed data k is taken to be:

${\displaystyle k=1/\left(\Phi ^{-1}(3/4)\right)\approx 1.4826}$

Therefore, one could improve the estimate of the single-sample Sharpe ratio by multiplying it by $\approx [.67,\, .8]$. Both answers seem like the right one, unless of course, one argues that sample moments are undefined for a single sample space.

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