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Suppose that $X_1, \ldots X_n$ are iid $N(\mu, \sigma^2)$ where $\mu$ and $\sigma$ are both unknown with $\mu \in \mathbb{R}$, $\sigma \in \mathbb{R}+$, $\boldsymbol{\theta} = (\mu, \sigma)$, $n \geq 2$. Consider the parametric function $\tau (\boldsymbol{\theta}) = e^{2(\mu + \sigma^2)}$. Derive the UMVUE for $\tau(\theta)$.

Solution:

We can show $\boldsymbol{T} = (\bar{X}_n, S^2_n)$ is complete sufficient, where $\bar{X}_n = \sum_{i=1}^{n} \frac{1}{n} X_i$ is the sample mean and $S^2_n = (n-1)^{-1} \sum_{i=1}^{n} (X_i - \bar{X})^2$ is the sample variance. I think best approach here is use $T = e^{2 X_1}$, so $E(T) = \tau(\theta)$, i.e., $T$ is unbiased. Then by the Lehmann-Scheffe Theorem I (link, pg. 371), the UMVUE is \begin{equation} E_{\theta}[e^{2 X_1} | \bar{X}_n, S^2_n] \end{equation}

Can we simplify this further? Is there another approach to try? Thanks.

Jackson
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  • Since mean and variance have different units, what does it mean to add them? – Dave May 01 '20 at 00:21
  • I'm not sure I follow the comment. Both the sample mean and sample variance are functions of the data. – Jackson May 01 '20 at 01:22
  • Mean in meters, variance in meters squared. Mean in dollars, variance in dollars squared... – Dave May 01 '20 at 01:31
  • I added the formulas for the sample mean and sample variance. You can add them directly. Both are functions of the data. – Jackson May 01 '20 at 02:17
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    There are inconsistencies in your question. The quantity you want to estimate in the title is different to the quantity stated in your question. Which is correct? – adriankahk May 01 '20 at 04:21
  • @Dave: since $\exp\{\mu+\sigma^2/2\}$ is the mean of a log-Normal $\mathcal{LN}(\mu,\sigma^2)$ variate, I do not see the issue in considering this quantity. – Xi'an May 01 '20 at 08:31
  • I've cleaned up the post. The quantity $\tau(\theta)$ in the question was correct, the title was incorrect. I also added the self-study tag. – Jackson May 01 '20 at 15:49
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    See https://stats.stackexchange.com/questions/105717. – whuber May 01 '20 at 17:25
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    @whuber, if I am following, it sounds like the Finney Function is the best hope to improve on my answer? – Jackson May 05 '20 at 01:46

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