6
$\begingroup$

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.

$\endgroup$
1

2 Answers 2

11
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Such a good answer. $\endgroup$
    – phdstudent
    Commented Jul 8, 2022 at 0:49
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.