# Prove bias/unbias-edness of mean/median estimators for lognormal

Looking at a problem where X is lognormally distributed from normal distribution Y, which asks me to prove that:

1) $$e^{\bar{y}}$$ is a biased estimator for the median of X

2) $$e^{\bar{y} - \sigma^2 / (2n)}$$ is an unbiased estimator for the median of X

3) $$e^{\bar{y} - \sigma^2 / (2n)}e^{\sigma^2/2}$$ is an unbiased estimator for $$\mu_x$$

I know that I'm being asked to solve for $$E(\hat\theta) = \theta$$, but I'm absolutely adrift as to how to actually calculate the expected value of the estimators. If someone could cluebat me, I would be appreciative.

• I presume you mean $X=\exp\{Y\}$ as the sentence "X is lognormally distributed from normal distribution Y" is unclear. Jun 7 '20 at 16:38

For example, for (1)

$$E[e^{\bar y}]=E\left[\prod_{i=1}^n e^{y_i/n}\right]=E[e^{y/n}]^n$$

Here, $$y$$ is a normal RV, and $$E[e^{y/n}]=E[e^{ty}]$$, where $$t=1/n$$ is the MGF, which is $$e^{\mu t + \sigma^2 t^2/2}$$. That said, $$E[e^{\bar y}]=(e^{\mu/n + \sigma^2/2n^2})^n=e^\mu e^{\sigma^2/2n}$$

Lognormal distribution's median is $$e^\mu$$, so the above expression is a biased estimator of the median of $$X$$.

• I understand the algebra and see how point 2 follows from point 1, but two points are unclear: 1) where does the product notation come from, as I would assume it to be a sum, and 2) wouldn't the expected value be the first moment/derivative of the MGF? Jun 7 '20 at 16:13
• 1) $e^{x+y}=e^xe^y$ 2) I'm just exploiting the similarity between the MGF of a normal RV and $E[e^{y/n}]$ to actually find the expected value. Jun 7 '20 at 16:18
• So am I missing some rule for manipulating summations? $\Sigma e^x = e^{x_0} + e^{x_1} ...$ which is not the same, according to my understanding Jun 7 '20 at 16:34
• $$e^{\bar y} = e^{\frac{1}{n}\sum_{i=1}^n y_i}=e^{\sum_{i=1}^n\frac{y_i}{n}}=\prod_{i=1}^n e^{y_i/n}$$ Jun 7 '20 at 16:36
• Okay, having worked through the problems, it's crystal-clear now. Thanks for helping me salvage something out of this semester. Jun 7 '20 at 17:30