0
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • $\begingroup$ 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, 2020 at 16:38

1 Answer 1

Reset to default
1
$\begingroup$

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 $X$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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, 2020 at 16:13
  • 1
    $\begingroup$ 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, 2020 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, 2020 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, 2020 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, 2020 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.