56
$\begingroup$

I have a basic question about annualized Sharpe Ratio Calculation: if I know the daily return of my portfolio, the thing I need to do is multiply the Sharpe Ratio by $\sqrt{252}$ to have it annualized.

I don't know why is that, can any body explain?

Improve this question
$\endgroup$
3
  • $\begingroup$ If I had daily data for one month (22 trading days), and wanted an estimate of annualized Sharpe ratio, would I use sqrt(252)? $\endgroup$ – user4850 Feb 26 '13 at 0:06
  • $\begingroup$ @GerryBlais Yes. $\endgroup$ – chrisaycock Feb 26 '13 at 0:53
  • 1
    $\begingroup$ @David : don't you think RYogi's answer could be accepted for your question? $\endgroup$ – SRKX Feb 26 '13 at 11:15
45
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ For me, this link led me to a page first asking me to prove I'm human, and then trying to install a chrome extension. $\endgroup$ – Eduardo Wada Apr 9 '19 at 10:42
  • $\begingroup$ Thanks @EduardoWada, I updated the link to a different source. $\endgroup$ – Ryogi Apr 9 '19 at 14:17
  • 1
    $\begingroup$ That Andy Lo paper "The Statistics of Sharpe Ratios" is awesome. The overall theme is how autocorrelations in high frequency (e.g. daily) returns complicate using them to compute a lower frequency (e.g. annual) Sharpe ratio. $\endgroup$ – HaroldFinch Feb 18 at 3:34
29
$\begingroup$

@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:

  1. The returns of the portfolio are a Wiener process, in which volatility scales with the square-root of time.
  2. There are 252 trading days in a year.

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

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Thanks, Tal. I think the assumption 1 should be the returns are IID. $\endgroup$ – David Oct 28 '11 at 16:04
17
$\begingroup$

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).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Hi Patrick, that's exactly what I have thought about. But why the annualized standard deviation should be sqrt(252) times its daily value. My calculation is stdev(252r)=252*stdev(r). $\endgroup$ – David Oct 28 '11 at 13:55
  • 3
    $\begingroup$ Correct: sd(252 * r) = 252 * sd(r) but that is not what is being done. You want sd( sum from i=1 to 252 of r_i ) $\endgroup$ – Patrick Burns Oct 28 '11 at 15:06
  • $\begingroup$ @PatrickBurns what's difference btn sd( sum from i=1 to 252 of r_i ) and sd(252*r) ? $\endgroup$ – user3595632 Dec 19 '18 at 0:49
17
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ How about some Latex? ;) $\endgroup$ – Ryogi Oct 29 '11 at 5:23
  • $\begingroup$ Ha I know, I was being lazy... $\endgroup$ – strimp099 Oct 29 '11 at 12:53
2
$\begingroup$

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'.

Improve this answer
$\endgroup$
2
$\begingroup$

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).

Improve this answer
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$
0
$\begingroup$

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:

  1. Calculate daily log returns:

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

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

  2. 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 $\mu - 0.5\sigma^2$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Interesting suggestion to reduce the mean return by the "volatility tax" when annualizing it for use in the Sharpe ratio. Do you have a good reference for doing this when calculating the Sharpe ratio? I think that you are correct: calculating an annual result from daily data requires the geometric mean return, not the arithmetic mean return. $\endgroup$ – HaroldFinch Feb 18 at 2:55
  • $\begingroup$ That said, did you make a sign error in your Sharpe ratio formula? I think that instead of $... + 0.5\sigma$ it should be $... - 0.5\sigma$ in order to be consistent with the "volatility tax" formula $\mu - 0.5\sigma^{2}$. Your + version means that the SR goes up more (or decreases less) as sigma goes up than the non-corrected result does, which seems to be incorrect. $\endgroup$ – HaroldFinch Feb 18 at 2:56
  • $\begingroup$ Another point: that "volatility tax" formula $\mu - 0.5\sigma^{2}$ is only exact for Geometric Brownian Motion, which is a continous process. When you have discrete returns, like daily returns, is that formula actually only the first 2 terms of a Taylor series expansion? $\endgroup$ – HaroldFinch Feb 18 at 3:00

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.