When the Square-Root Rule Annualizes a Sharpe Ratio

# How to annualize Sharpe Ratio?


# How to annualize Sharpe Ratio?












If I know the daily returns of my portfolio, I need to multiply the Sharpe Ratio by $\sqrt{252}$ to have it annualized. I don't understand why that is.

## Answer by Ryogi (score 50)

https://quant.stackexchange.com/a/2261

Actually, that is not always the case. Here is a great paper by Andy Lo, "The Statistics of Sharpe Ratios". He shows how monthly Sharpe ratios "cannot be annualized by multiplying by $\sqrt{12}$ except under very special circumstances". I expect this will carry over to annualizing daily Sharpe Ratios.

## Answer by Tal Fishman (score 35)

https://quant.stackexchange.com/a/2262

@RYogi's answer is definitely far more comprehensive, but if you're looking for what the assumptions behind the common rule of thumb are, they are:

- The returns of the portfolio are a Wiener process, in which volatility scales with the square-root of time.

- There are 252 trading days in a year.

As Lo's paper points out, assumption #1 is somewhat suspect.

## Answer by strimp099 (score 20)

https://quant.stackexchange.com/a/2271

You often see various financial metrics scale with the square root of time This stems from the process that drives the lognornmal returns in stock prices which is the Ito process $dS = \mu Sdt + \sigma SdZ$.

The Wiener process assumes that each dt is IID and has constant $\mu$ and $\sigma^2$, therefore the same expected value and variance at each increment. Because: $$\operatorname{Var}\left(\ln\left(\frac{S(T)}{S(T_0)} \right) \right) = \operatorname{Var}\left(\ln\left(\frac{S(t_n)}{S(t_{n-1})} \right) \right) + \operatorname{Var}\left(\ln\left(\frac{S(t_{n-1})}{S(t_{n-2})} \right) \right) + \dots + \operatorname{Var}\left(\ln\left(\frac{S(t_1)}{S(t_0)} \right) \right)$$ $$ = ns^2 = s^2\frac{(T-T_0)}{dt}$$ $$\text{where } n = \frac{(T-T_0)}{dt} $$ $$\text{where } S(T_{n})=S_{0}e^{(u-\frac{1}{2}\sigma^2)T_n+\sigma W(T_n)}$$ It follows that $s^2(T-T_0)$. Because the variance should be finite, in the $\lim_{dt \rightarrow 0}$, variance should be proportional to $dt$. Since $s^2$ is 1 for a lognormally distributed process, the variance is $(T-T_0)$, the standard deviation is therefore $\sqrt{T-T_0}$ or $\sqrt{T}$.

The reason you see financial metrics scaled to the square root of time is because the metrics are usually calculated using stock returns, which are assumed to be lognormally distributed. Whether it's right or wrong really has to do with your assumption of how stock returns are really distributed.

## Answer by Patrick Burns (score 19)

https://quant.stackexchange.com/a/2270

Here's the idea of where that comes from:

To annualize the daily return, you multiply by 252 (the number of observations in a year).

To annualize the variance, you multiply by 252 because you are assuming the returns are uncorrelated with each other and the log return over a year is the sum of the daily log returns.

So the annualization of the ratio is 252 / sqrt(252) = sqrt(252).

## Answer by Stephen P (score 3)

https://quant.stackexchange.com/a/27712

The reason is that the Sharpe Ratio is typically defined in terms of annual return and annual deviation. As everyone has said, you go from daily returns to annual returns by assuming daily returns are independent and identically distributed.

With that assumption, you get annual return by multiplying by daily return by 252 (compounding makes little difference when daily return is 1 bp). You get annual deviation by multiplying daily deviation by square root of 252. So you get the annual ratio by the daily ratio by 252/sqrt(252) = sqrt(252).

## Answer by shabbychef (score 2)

https://quant.stackexchange.com/a/3675

The units of Sharpe ratio are 'per square root time', that is, if you measure the mean and standard deviation based on trading days, the units are 'per square root (trading) day'. It should be obvious then, how to re-express Sharpe ratio in different units. For example, to get to 'per root month', multiply by $\sqrt{253/12}$.

The reason why Sharpe has these units is because the drift term has units of 'return per time', while variance is 'returns squared per time'.

## Answer by Magic is in the chain (score 1)

https://quant.stackexchange.com/a/41438

It is not that hard at all. You making two assumptions:

1) There are 252 business days in a year. 2) The daily returns are IID (independently and identically distributed).

Then:

a) the sum of the means of the 252 IID will be 252 times their identical mean. b) the sum of their variances will be 252 times their identical variance, as independence implies their cov are zero. Standard deviation then, by definition, would be square root of 252 times their std dev.

As Sharpe ratio is mean over std dev, you get the square root of 252 as the scaling factor for converting ratio computed on daily returns into its annualised equivalent.

## Answer by Konstantin (score 0)

https://quant.stackexchange.com/a/34596

If you assume that your capital $C$ can be described as a Geometric Brownian Motion (GBM) with annualized drift $\mu$ and annualized standard deviation $\sigma$, then a correct way to calculate Sharpe Ratio ($\frac{\mu}{\sigma}$, disregarding risk free rate) would be as follows:

- Calculate daily log returns:

$$\log(C_{t+1}) - \log(C_{t})$$

- Find mean ($\bar{\mu}$) and standard deviation ($\bar{\sigma}$) of the daily log returns

- Calculate Sharpe:

$$\frac{\mu}{\sigma} = \left(\frac{\bar{\mu}}{\bar{\sigma}} + 0.5 \bar{\sigma}\right) \times \sqrt{DaysPerYear}$$

The reason to add $0.5 \bar{\sigma}$ to the log return-based ratio can be explained via Ito's lemma: logarithm of a GBM has annualized expectation of $\mu - 0.5\sigma^2$.