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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.

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

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  • $\begingroup$ 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? $\endgroup$
    – tbert
    Jun 7 '20 at 16:13
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    $\begingroup$ 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. $\endgroup$
    – gunes
    Jun 7 '20 at 16:18
  • $\begingroup$ 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 $\endgroup$
    – tbert
    Jun 7 '20 at 16:34
  • $\begingroup$ $$e^{\bar y} = e^{\frac{1}{n}\sum_{i=1}^n y_i}=e^{\sum_{i=1}^n\frac{y_i}{n}}=\prod_{i=1}^n e^{y_i/n}$$ $\endgroup$
    – gunes
    Jun 7 '20 at 16:36
  • $\begingroup$ Okay, having worked through the problems, it's crystal-clear now. Thanks for helping me salvage something out of this semester. $\endgroup$
    – tbert
    Jun 7 '20 at 17:30

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