Why the Sharpe Ratio Uses Standard Deviation and Its Scaling Limits

# Logic behind sharpe ratio


# Logic behind sharpe ratio












I have a confusion regarding how the sharpe ratio is derived. My question is why the denominator contains the standard deviation of returns of portfolio? I mean why did someone came to this conclusion and how did he eliminated the possibilities that it isn't raised to some power?

## Answer by Raskolnikov (score 3, accepted)

https://quant.stackexchange.com/a/37242

What would it mean to compare return with a power of the standard deviation different from 1? Note that the standard deviation is expressed in the same units as the return. Let's say you have a return in dollars, then the standard deviation is also in dollars. But the variation is in dollars squared, not to speak about other powers.

But also think about distributions, in particular the normal distribution. You surely learned the rule that 68% of returns are between "mean return-standard deviation" and "mean return+standard deviation", 95% are between "mean return-2*standard deviation" and "mean return+2*standard deviation", etc... Standard deviation is the natural measure for your spread on the possible values w.r.t. the mean return.

So it is natural to consider a ratio of return to standard deviation and to try to maximise it. It's interpreted as maximizing the return w.r.t. risk.

## Answer by vonjd (score 7)

https://quant.stackexchange.com/a/37250

Another intuitive interpretation of the Sharpe ratio is as a signal-to-noise ratio: $$\frac{\mu}{\sigma}$$ where you compare the strength of the signal (= return) to the level of noise (= risk).

The bigger this ratio is the better: either you have more return (= signal) or you have less risk (= noise).

## Answer by Matthew Gunn (score 5)

https://quant.stackexchange.com/a/37277

People compute the Sharpe ratio because it has some useful properties.

#### If you increase the leverage of a strategy, the Sharpe ratio remains the same

Let $R$ be the return to some strategy. And let $R_f$ be the risk free rate. The Sharpe Ratio of return $R$ is given by:

$$ \frac{\operatorname{E}[R - R_f]}{\sigma(R - R_f)} $$

Now imagine that we executed the strategy but with $5 \times$ leverage. What's the Sharpe Ratio?

$$ \frac{\operatorname{E}[5R - 5R_f]}{\sigma(5R - 5R_f)} =\frac{5\operatorname{E}[R - R_f]}{5\sigma(R - R_f)} = \frac{\operatorname{E}[R - R_f]}{\sigma(R - R_f)} $$

It's the same thing! Both the expected excess return and the standard deviation of the excess return increase linearly in leverage, hence the ratio remains the same.

This is a nice property. You can scale up the expected return of a strategy by increasing leverage, but you can't raise the Sharpe ratio.

On the other hand...

#### The Sharpe ratio is NOT invariant to time horizon

If you switch between daily, monthly, or yearly returns, the Sharpe ratio will change! Corollary: to sensibly compare Sharpe ratios, everything must be in the same units of time.

For simplicity, let's assume each time period is independent and identically distributed (IID). Let $R^d- R_f^d$ be the excess return of a single day. The Sharpe ratio for the daily frequency would be:

$$ \frac{\operatorname{E}[R^d- R_f^d]}{\sigma [R^d- R_f^d]}$$

If we had a monthly return $R^m$ that's the sum of 21 IID daily returns, we would have:

- $\operatorname{E}[R^m- R_f^m] = 21 \operatorname{E}[R^d- R_f^d]$

- $\sigma^2 [R^m- R_f^m] = 21 \sigma^2 [R^d- R_f^d]$

- $\sigma [R^m- R_f^m] = \sqrt{21} \sigma [R^d- R_f^d]$

Hence $ \frac{\operatorname{E}[R^m- R_f^m]}{\sigma [R^m- R_f^m]} = \sqrt{21} \frac{\operatorname{E}[R^d- R_f^d]}{\sigma [R^d- R_f^d]} $.

On the other hand, the ratio $\frac{\mu}{\sigma^2}$, expected return divided by variance, for an excess return is invariant to time horizon (but not invariant to leverage). As @noob2 pointed out, there are numerous formulas in finance where $\frac{\mu}{\sigma^2}$ comes out in the math.

## Answer by RobAbMo (score 2)

https://quant.stackexchange.com/a/37251

To echo, @Raskolnikov, this is about dimensions/units.

The Sharpe ratio is a measure of expected excess return per unit of risk. In order for this to be a normalised quantity that can compare across different assets/portfolios/etc... it's best for it to be a dimensionless ratio.

Therefore we use standard deviation since is a useful quantity that fulfils the requirement and is easily calculated.

I agree that it isn't the only measure we could use, we could calculate a variability (or volatility as we speak of in finance generally) of returns using a different measure (i.e. different assumptions about the underlying distribution of returns). However, the Sharpe ratio (and the rest of the family of similar measures) has evolved to be the standard and therefore the measure for which most practitioners have a good intuition.

## Answer by Dave Harris (score 0)

https://quant.stackexchange.com/a/37302

Look at the Sharpe ratio again. It is derived under the assumption that all parameters are known. As such, it is a Frequentist tool. It is not actually restricted so that the risk-free rate is the reference rate. Technically, any rate could be the reference rate. It is an expected distance scaled for natural variability. That is to say, it is the z-test. Of course, since the variance is not known, it is the t-test in practice.

It is a pivotal quantity of the Gaussian, which is why there is no other power or anything else in there. The difficulty is that there exists a proof that this method cannot work unless the parameters are actually known with probability one.

The method falls apart because the basic equation of profitability of an investment has no non-Bayesian solution. That is to say $$w_{t+1}=Rw_t+\epsilon_{t+1}, R>1,$$ only has a perfectly inefficient estimator of the mean and via the mean, the variance is also perfectly imprecise. Simplified, as the sample size goes to infinity, the precision of the estimates goes to zero. It creates the seeming paradox that the Sharpe ratio is only accurate up to sample sizes of two.

This is not well known in the literature.

As interestingly, the Bayesian estimator does not at all look like the Frequentist estimator. They clearly do not point to the same value. However, by the Wald Complete Class Theorem, it is necessarily true that the Frequentist estimator is an inadmissible statistic. The Frequentist solution, the Sharpe ratio, is outside the complete class of valid solutions, unless it is true that all users really do know the true values of the parameters.