Why geometric returns fall short of arithmetic returns

Question

An asset returns a random simple return $R$ each period with mean $\mathbb{E}[R] = \mu$ and variance $\sigma^2$ . Compounding cares about the log growth factor $\log(1 + R)$ , and the long-run growth rate is driven by $\mathbb{E}[\log(1 + R)]$ .

How does $\mathbb{E}[\log(1 + R)]$ compare to $\log(1 + \mu)$ ? Which way is the gap, and how big is it?

Accepted Answer

Direction, from Jensen. The function g(x) = \log(1 + x) has g''(x) = -1/(1+x)^2 < 0, so it is strictly concave. Jensen's inequality for concave g gives \mathbb{E}[g(R)] \le g(\mathbb{E}[R]), so \mathbb{E}[\log(1 + R)] \ \le \ \log(1 + \mu). The expected log growth is below the log of the expected return: an investor's realized compounded growth rate falls short of what the average return alone would suggest. Equality only if R has no variance. Size, from the delta method. Expand around \mu: \mathbb{E}[\log(1 + R)] \approx \log(1 + \mu) + 	frac12 g''(\mu)\,\sigma^2 = \log(1 + \mu) - \frac{\si…

Why geometric returns fall short of arithmetic returns

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