Mathematical proof of $g = \mu - \frac{\sigma^2}{2}$ relationship between CAGR and average returns

I found in a paper the relation between the CAGR and the arithmetic average of returns to be

$$g \sim \mu - \frac{\sigma^2}{2}$$

where g is the geometric average, $$\mu$$ the arithmetic average and $$\sigma^2$$ the variance of the returns. I cannot find any formal derivation of this relationship.

Here is a straightforward derivation of this approximate relationship for a discrete sample of returns that does not require specifying an underlying probability distribution or a continuous stochastic process (e.g., geometric Brownian motion).

Given a finite sequence of returns $$r_1,r_2, \ldots, r_n$$, the CAGR $$g$$ and arithmetic average return $$\mu$$ are $$g = \left[\prod_{j=1}^n(1+r_j)\right]^{1/n} -1, \quad \mu = \frac{1}{n}\sum_{j=1}^n r_j$$

Rearranging the CAGR formula and taking logarithms, we get
$$\tag{1}\log(1+g) = \frac{1}{n}\sum_{j=1}^n \log(1+r_j)$$

By expanding $$\log(1+r_j)$$ around $$\log(1+\mu)$$ and substituting into (1) we can obtain the relationship between $$g$$ and $$\mu$$. Note that

$$\tag{2}\log(1+r_j) - \log(1+\mu) = \log\left(\frac{1+r_j}{1+\mu}\right) = \log \left(1 + \frac{r_j-\mu}{1+\mu} \right)$$

The Taylor series $$\log(1+\lambda_j) = \lambda_j - \lambda_j^2/2 + \lambda_j^3/3 \mp \ldots$$ converges for $$|\lambda_j| < 1$$. Applying this to (2) with $$\lambda_j = (r_j- \mu)/(1+\mu)$$, we get

$$\tag{3}\log(1+r_j) = \log(1+\mu) + \lambda_j - \frac{\lambda_j^2}{2} + \mathcal{O}(\lambda_j^3)$$

Substituting into (1) with (3) yields

$$\tag{4}\log(1+g) = \log(1+\mu) + \frac{1}{n}\sum_{j=1}^n \lambda_j - \frac{1}{2n}\sum_{j=1}^n \lambda_j^2 + \frac{1}{3n}\mathcal{O}\left(\sum_{j=1}^n \lambda_j^3\right)$$

The second term on the RHS must vanish since $$\sum_{j=1}^n\lambda_j = \sum_{j=1}^n(r_j -\mu) = n\mu - n\mu$$. The third term is just $$\sigma^2/(2(1+\mu)^2)$$ where the volatility estimator $$\sigma$$ is given by

$$\sigma^2:= \frac{1}{n}\sum_{j=1}^n (r_j- \mu)^2$$

We can also assume that the error term $$\frac{1}{3n}\mathcal{O}\left(\sum_{j=1}^n \lambda_j^3\right)$$ can be neglected for typical (small) values of $$r_j$$.

Making substitutions and applying the exponential function to both sides of (5) it follows that

$$\tag{6}1 + g \approx (1+\mu)\exp \left(- \frac{\sigma^2}{2(1+\mu)^2}\right)$$

Finally, expanding the exponential function and using the approximation $$\exp(-x) \approx 1-x$$ we get

$$1+g \approx (1 + \mu)\left[1-\frac{\sigma^2}{2(1+\mu)^2}\right] =1 +\mu - \frac{\sigma^2}{2(1+\mu)},$$

and, hence,

$$g \approx \mu - \frac{\sigma^2}{2(1+\mu)} \approx \mu - \frac{\sigma^2}{2},$$

where an additional approximation $$1/(1+\mu) \approx 1$$ is used.

For typical indices like the S&P 500, the error of this approximation is a few basis points.

• Such a good answer. Commented Jul 8, 2022 at 0:49

It's a special case of the AM-GM inequality, assuming that market returns follow a lognormal distribution.

Consider the simple example of a stock that has a 50% probability of rising and falling 10% every period.

Its arithmetic average is obviously 0: (50% * +10%) + (50% * -10%) = 0

Its geometric average is (1+10%)^0.5*(1-10%)^0.5 -1 = -0.5%

Or a more extreme example the other way. A stock that doubles and halves with equal probability. It's geometric average is obviously 0. In the long-run, there's a doubling for every halving, and vice versa. But the arithmetic average is (50% * +100%) + (50% * -50%) = +25%.

For the full continuous lognormal distribution rather than my discrete examples above, the half variance formula above is derived from calculating its first moment. http://mathworld.wolfram.com/LogNormalDistribution.html