Four Years of Risk Is Twenty Percent, Not Forty

The standard deviation of a sum is not the sum of the standard deviations, so quadrupling the horizon only doubles the risk. The article carries the general square-root law, the ratio that diagnoses the mistake, and the controls showing a bell curve does none of the work: a two-point yearly return lands on 0.20026 and a uniform one on 0.20008. It also carries what actually breaks the rule, which is dependence rather than fat tails.

Yearly returns on some instrument have a standard deviation of ten percent. Hold it for four years instead of one. How wide is the spread of the four-year return?

Forty percent is the answer almost everyone gives first, and it is wrong by a factor of two. The correct figure is twenty percent. Quadrupling the holding period only doubles the risk, and the reason is a single line about which quantity is the one that adds.

Why forty percent feels right

The reflex is a proportionality argument, and proportionality arguments are usually a good instinct. Four years, four times the exposure, four times the spread. It even survives a sanity check of the units, because a standard deviation and a return are measured in the same thing.

What it skips is that the standard deviation of a sum is not the sum of the standard deviations. Nothing about risk is additive except the square.

The one quantity that adds

Continuously compounded returns

Write $R_t = \ln(S_t / S_{t-1})$ for the log return of year $t$ . Then the return over the whole four years is $R_1 + R_2 + R_3 + R_4$ exactly, with no cross terms, because logarithms turn the product of four growth factors into a sum. That additivity is the only thing continuous compounding is buying here.

Assume the four yearly returns are independent and identically distributed. For independent terms the variance of a sum really is the sum of the variances, so:

\operatorname{Var}\!\left(\sum_{t=1}^{n} R_t\right) = \sum_{t=1}^{n} \operatorname{Var}(R_t) = n\,\sigma^2

(1)

A one-year standard deviation of $\sigma = 1/10$ means a one-year variance of $1/100$ . Four years carry four copies of it, which is $4/100 = 1/25$ . Take the square root of that and you are back in the units of a return:

\sigma_4 = \sqrt{4 \times \tfrac{1}{100}} = \sqrt{\tfrac{1}{25}} = \tfrac{1}{5} = 20\%

(2)

Fig. 1 — The square of the spread climbs in equal steps. The spread itself climbs as a root, so it is already most of the way there after one year.

The general law, and the number two

Nothing above depended on the horizon being four years or on the yearly figure being ten percent. In general:

\sigma_n = \sqrt{n}\,\sigma_1

(3)

Multiply any horizon by four and the spread multiplies by exactly $\sqrt{4} = 2$ , whatever the horizon was and whatever the one-period figure was. That is worth remembering as a ratio rather than as an answer, because it also diagnoses the mistake: the wrong answer and the right one differ by exactly the factor $\sqrt{4}$ , which is a signature. If someone hands you a risk figure that is off by $\sqrt{n}$ , they scaled the spread where they should have scaled its square.

The same equation is what lets a desk quote a daily figure and an annual figure as the same object. With roughly 252 trading days in a year, an annual twenty percent corresponds to a daily $0.20/\sqrt{252} \approx 1.26\%$ . Read equation (3) in that direction and it is a conversion; read it forward and it is the answer to the question above.

Fig. 2 — The two answers are the same at one year and separate afterwards. The gap at four years is a factor of two.

One consequence is worth pulling out, because it is where the square root earns its keep. If the yearly return also has a mean $\mu$ , the mean over $n$ years grows like $n\mu$ while the spread grows only like $\sqrt{n}\,\sigma$ . Their ratio therefore grows like $\sqrt{n}$ :

\frac{n\mu}{\sqrt{n}\,\sigma} = \sqrt{n}\,\frac{\mu}{\sigma}

(4)

Any edge you have becomes more visible relative to the noise the longer you hold it, at a rate of the square root of the horizon and no faster. That single line is the reason time diversifies and the reason it does so slowly.

The bell curve is not doing the work

Root-time scaling gets taught alongside Brownian motion often enough that people file it as a property of the normal distribution. It is not. Equation (1) needs the four yearly returns to be independent with a finite variance, and it needs nothing else. No symmetry, no bell shape, no continuity.

A quick way to see that is to replace the yearly return with a coin flip that pays plus or minus ten percent, which is about as far from a bell curve as a random variable gets. All sixteen four-year paths can be enumerated by hand, and the four-year variance comes out to exactly $1/25$ , with no sampling involved at all. A uniform yearly return with the same standard deviation lands in the same place. Both were checked here against the exact figure of 0.20000: the two-point law measured 0.20026 and the uniform law 0.20008 over two hundred thousand sampled paths each, with a normal control at 0.20068.

This matters more than it looks. Someone who believes the square root comes from normality will abandon the formula the moment returns look fat-tailed, and that is the wrong lesson. The formula survives fat tails. It does not survive dependence.

Where the square root stops holding

Independence is load-bearing, so drop it and watch the answer move. Suppose consecutive yearly returns are correlated with coefficient $\rho$ . The variance of the sum picks up every covariance term:

\operatorname{Var}\!\left(\sum_{t=1}^{n} R_t\right) = n\sigma^2 + 2\sigma^2\!\!\sum_{1 \le s < t \le n}\!\! \rho_{\,t-s}

(5)

With positive autocorrelation, which is what a trending market looks like, the four-year figure comes out above twenty percent, and a desk that annualises by $\sqrt{n}$ is understating its risk. With negative autocorrelation, the mean-reverting case, four years are calmer than the root suggests. The exponent on $n$ stops being one half in both directions, and long-memory processes are usually described by exactly that: an exponent other than $1/2$ .

Two smaller conditions are worth naming. The variance has to be finite, which rules out laws with tails heavy enough that $\sigma^2$ does not exist. And the returns have to be the additive kind. Simple returns compound multiplicatively, so their variances do not add cleanly, and the mismatch between the two conventions is where a second factor of confusion usually enters. Over one year at these volatilities the two agree closely enough that nobody notices, which is precisely why it goes unnoticed.

Sources and further reading

Variance, and its additivity over independent terms — Variance
The scaling rule in its finance form — Volatility (finance)
Where the root-time picture comes from, and what it assumes — Random walk
The exponent that replaces one half under long memory — Hurst exponent

Every figure above was checked three ways: symbolically, by exhaustive enumeration of all sixteen paths of a two-point yearly return in exact rational arithmetic, and by simulation under three different yearly laws at horizons of 1, 2, 3, 4, 9, 16 and 25 years.