2.2 Volatility drag: arithmetic versus geometric growth

Motivation

The return you can expect on one day and the rate at which your wealth compounds over years are different numbers. The difference is about $\sigma^2/2$ per year, which for a 30%-volatility small-cap portfolio is 4.5 percentage points a year. It is the single most important fact for position sizing: M03's Kelly criterion is this subsection with leverage added.

The growth rate of wealth#

Setup

Let one-period gross returns $1 + R_t$ be iid, positive, with $E[R_t] = \mu$ , $\operatorname{Var}(R_t) = \sigma^2$ . Wealth after $T$ periods is $W_T = W_0 \prod_{t=1}^{T} (1 + R_t)$ .

Proposition 2.2Jensen bound and second-order drag

$E[\ln(1+R)] \le \ln(1+\mu)$ , with equality iff $R$ is degenerate. If $R$ has a finite third absolute moment and is small,

E[\ln(1+R)] = \ln(1+\mu) - \frac{\sigma^2}{2(1+\mu)^2} + O\!\left(E|R-\mu|^3\right) \approx \mu - \frac{\sigma^2}{2}. \tag{2.3}

Proof

$\ln$ is strictly concave, so Jensen gives the inequality, strict unless $R$ is a.s. constant. For the expansion, Taylor-expand $\ln(1+x)$ about $x = \mu$ with Lagrange remainder:

\ln(1+R) = \ln(1+\mu) + \frac{R-\mu}{1+\mu} - \frac{(R-\mu)^2}{2(1+\mu)^2} + \frac{(R-\mu)^3}{3(1+\xi)^3}

for some $\xi$ between $\mu$ and $R$ . Take expectations: the linear term vanishes, the quadratic term gives $-\sigma^2/(2(1+\mu)^2)$ , and the remainder is bounded by a constant times $E|R-\mu|^3$ when $1 + \xi$ is bounded away from 0. The final approximation uses $\ln(1+\mu) \approx \mu$ and $(1+\mu)^2 \approx 1$ for daily or monthly $\mu$ .∎

By the strong law of large numbers, $\frac1T \ln(W_T/W_0) = \frac1T \sum_t \ln(1+R_t) \to E\ln(1+R)$ almost surely. So the long-run compound growth rate is $E\ln(1+R) \approx \mu - \sigma^2/2$ , not $\mu$ . The term $\sigma^2/2$ is the volatility drag.

Mean versus median under GBM#

Proposition 2.3exactness under lognormality

If $\ln(1+R) \sim N(m, s^2)$ , then $E[1+R] = e^{m + s^2/2}$ , so the expected log return is exactly $m = \ln(1+\mu) - s^2/2$ . In particular, for geometric Brownian motion $dS_t = \mu S_t\,dt + \sigma S_t\,dW_t$ ,

\ln\frac{S_T}{S_0} \sim N\!\left((\mu - \tfrac12\sigma^2)T,\ \sigma^2 T\right), \quad E[S_T] = S_0 e^{\mu T}, \quad \operatorname{median}(S_T) = S_0 e^{(\mu - \sigma^2/2)T}. \tag{2.4}

Proof

The moment-generating function of $X \sim N(m, s^2)$ at 1 is $E e^{X} = e^{m + s^2/2}$ ; take logs. For GBM, the solution $S_T = S_0 \exp((\mu - \sigma^2/2)T + \sigma W_T)$ follows from Itô's lemma applied to $\ln S_t$ (derived in M08 · 8.5). Given that solution, the mean follows from the MGF with $m = (\mu - \sigma^2/2)T$ , $s^2 = \sigma^2 T$ , and the median of a lognormal is $e^m$ because $\exp$ is monotone.∎

Corollary 2.3a

Under GBM, $P(S_T < E[S_T]) = \Phi(\sigma\sqrt{T}/2) \to 1$ as $T \to \infty$ .

Proof

$S_T < S_0 e^{\mu T}$ iff $(\mu - \sigma^2/2)T + \sigma W_T < \mu T$ iff $W_T/\sqrt T < \sigma\sqrt T/2$ .∎

For $\sigma = 20\%$ and $T = 20$ years, $\Phi(0.447) = 0.67$ : two-thirds of outcomes end below the mean. The mean is pulled up by a few very good paths. A "typical" investor experiences the median.

Arithmetic and geometric means#

Proposition 2.4AM–GM gap

For realised returns $R_1, \dots, R_T > -1$ , define the arithmetic mean $A = \frac1T \sum R_t$ and the geometric mean $G = \left(\prod (1+R_t)\right)^{1/T} - 1$ . Then $G \le A$ , with equality iff all $R_t$ are equal, and

\ln(1+G) \approx \ln(1+A) - \tfrac12 s^2, \qquad s^2 = \tfrac1T \textstyle\sum (R_t - A)^2. \tag{2.5}

Proof

AM–GM applied to the positive numbers $1 + R_t$ gives $(\prod(1+R_t))^{1/T} \le \frac1T \sum (1+R_t) = 1 + A$ , with equality iff all are equal. For (2.5), $\ln(1+G) = \frac1T \sum \ln(1+R_t)$ ; apply the expansion in Proposition 2.2 to the empirical distribution of $\{R_t\}$ , whose mean is $A$ and variance $s^2$ .∎

Worked example 2.2.1Nifty sample

From 17 Sep 2007 (close 4,494.65) to 4 Nov 2026 (close 23,960.60), 19.13 calendar years, the mean annual log return of the Nifty price index was $\ln(23{,}960.60/4{,}494.65)/19.13 = 8.75\%$ , a compound annual growth rate (CAGR) of $e^{0.0875} - 1 = 9.14\%$ . Annualised volatility was $20.2\%$ . Proposition 2.2 then says the arithmetic mean annual return was about $8.75\% + 0.202^2/2 = 10.8\%$ . A brochure quoting "average annual return 10.8%" and one quoting "CAGR 9.1%" describe the same index.

Worked example 2.2.2same mean, different growth

Two funds both have $\mu = 10\%$ . Fund A has $\sigma = 15\%$ , Fund B (a concentrated small-cap fund) has $\sigma = 40\%$ . Growth rates: $0.10 - 0.0113 = 8.9\%$ versus $0.10 - 0.08 = 2.0\%$ . Over 20 years the median outcome per ₹1 lakh invested is $e^{1.775} = \text{₹}5.9$ lakh versus $e^{0.40} = \text{₹}1.5$ lakh. Volatility is not just "risk"; at fixed arithmetic mean it directly lowers compound growth.

Practitioner notes

Leveraged and inverse ETFs that rebalance daily suffer drag proportional to $L^2\sigma^2/2$ (M03 · 3.4 makes this precise). Holding a 2× product for months does not give twice the index's return.
Mutual-fund factsheets in India report CAGR (and XIRR for SIPs). Academic papers often report arithmetic means. Convert before comparing.
In forecasting, the expected arithmetic return is the right input for one-period portfolio optimisation (M14), while the geometric return is the right input for long-horizon wealth projections. Mixing them up overstates retirement corpus projections by roughly $\sigma^2/2$ per year compounded.