Estimating a Sharpe Ratio from a Single Return

# Calculate Sharpe ratio for only one return


# Calculate Sharpe ratio for only one return












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?

## Answer by David Addison (score 3, accepted)

https://quant.stackexchange.com/a/33634

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.