Is the exponential of an unbiased mean unbiased?

Question

You estimate the mean log-return \mu with the sample mean \bar X (unbiased), then report a growth factor by exponentiating: \hat G = e^{\bar X} as an estimate of e^{\mu}. Assume the returns are Normal so \bar X \sim N(\mu, \sigma^2/n). Is e^{\bar X} an unbiased estimator of e^{\mu}? If not, which way is it biased, and by how much?

Accepted Answer

Exponentiating breaks unbiasedness. Unbiasedness gives \mathbb{E}[\bar X] = \mu, but the exponential is nonlinear, so \mathbb{E}[e^{\bar X}] 
e e^{\mathbb{E}[\bar X]} in general. Jensen gives the direction. The exponential g(x) = e^{x} is strictly convex, so \mathbb{E}[e^{\bar X}] e^{\mathbb{E}[\bar X]} = e^{\mu}. The estimate is biased high. The exact size (normal case). For a normal \bar X \sim N(\mu, \sigma^2/n), the moment generating function gives the exact expectation: \mathbb{E}[e^{\bar X}] = e^{\mu + \frac{1}{2}\cdot\frac{\sigma^2}{n}} = e^{\mu}\cdot e^{\sigma^2/(2n)}. So the growth …

Is the exponential of an unbiased mean unbiased?

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