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Assume that

$Y \sim N(\mu,\sigma^2)$

$X = e^Y$

Then X is lognormal distributed with parameters $\sigma$ and $\mu$.

I know that

$E(X) = E(e^Y) = \eta e^{\frac{\sigma^2}{2}} $

The median in the lognormal distrubution is $\eta = e^\mu$, and that $\eta* = e^\widehat{\mu}$ is not an unbiased estimator.

$X_1,...,X_n$ is independent and lognormal distributed. $Y_i = ln(X_i)$ for $i = 1,...,n$. We assume that we know the value of $\sigma$.

$\widehat{\mu} = \bar Y = \frac{1}{n}\sum_{i=1}^n Y_i = \frac{1}{n}\sum_{i=1}^n ln(X_i) $

Now I have to show that $\widehat{\eta} = e^{\widehat{\mu}-\frac{\sigma^2}{2n}} $ is an unbiased estimator for the median, and I don't know how to do this.

Any help would be greatly appreciated!

Joe6969
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1 Answers1

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First find that:$$\mathbb{E}X^{\alpha}=e^{\alpha\mu+\frac{1}{2}\alpha^{2}\sigma^{2}}$$

For that have a look at this answer.

Then:

$$e^{\hat{\mu}}=e^{\frac{1}{n}\sum_{i=1}^{n}\ln X_{i}}=\prod_{i=1}^{n}X_{i}^{\frac{1}{n}}$$

so that: $$\mathbb{E}e^{\hat{\mu}}=\prod_{i=1}^{n}\mathbb{E}X_{i}^{\frac{1}{n}}=\left(\mathbb{E}X_{1}^{\frac{1}{n}}\right)^{n}=\left(e^{\frac{1}{n}\mu+\frac{1}{2n^{2}}\sigma^{2}}\right)^{n}=e^{\mu+\frac{1}{2n}\sigma^{2}}$$

Then consequently:$$\mathbb{E}e^{\hat{\eta}}=\mathbb{E}e^{\hat{\mu}-\frac{1}{2n}\sigma^{2}}=e^{-\frac{1}{2n}\sigma^{2}}\mathbb{E}e^{\hat{\mu}}=e^{-\frac{1}{2n}\sigma^{2}}e^{\mu+\frac{1}{2n}\sigma^{2}}=e^{\mu}=\eta$$

drhab
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