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Let $X_1, X_2, \dots,X_n$ be the random samples from $f(x,\theta)$ = $\frac{2x}{\theta^2}$, $0 < x < \theta$, $\theta > 0$, then find the Uniformly minimum variance unbiased estimator of $\theta^2$.

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  • $\begingroup$ Do you what you have to find? $\endgroup$ – Davide Giraudo Feb 11 '12 at 16:09
  • $\begingroup$ Do you know that there is a website like this one for stats questions? I think the name is "cross-validated". $\endgroup$ – Gerry Myerson Feb 12 '12 at 1:11
  • $\begingroup$ @GerryMyerson And the site link is Cross Validated(stats.stackexchange.com) $\endgroup$ – Sasha Feb 12 '12 at 3:14
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The likelihood on the sample is $$ \mathcal{L}(x_1,\ldots,x_n) = \frac{2^n}{\theta^{2n}} \left( \prod_{k=1}^n x_k \right)\left[ \min(x_1,\ldots,x_n) > 0\right] \cdot \left[ \max(x_1,\ldots,x_n) < \theta \right] $$ where $\left[ \bullet \right]$ is the Iverson bracket.

The function $\theta^{-2 n}$ is monotonically decreasing function, hence the maximum of the likelihood occurs at $\theta = \max(x_1,x_2, \ldots, x_n)$. Thus $T=\max(X_1,X_2, \ldots, X_n) = X_{n:n}$ is the complete sufficient statistics, as a maximum likelihood estimator.

It is easy to see that $\delta(x_1,\ldots,x_n) = \frac{2}{n} \left( x_1^2 + x_2^2 + \cdots + x_n^2 \right)$ is an unbiased estimator for $\theta^2$ as $$ \mathbb{E}(\delta(X_1,X_2,\ldots,X_n)) = \mathbb{E}(2 X^2) = \theta^2 $$ The MVUE for $\theta^2$ is thus $$ \begin{eqnarray} \eta\left( X_1,\ldots,X_n\right) &=& \mathbb{E}\left( \delta(X_1,\ldots,X_n) | T \right) \\ &=& \mathbb{E}\left( \frac{2}{n} \sum_{k=1}^n X_{k:n}^2 | T \right) \\&=& \frac{2}{n} \sum_{k=1}^{n-1} X_{k:n}^2 + \frac{2}{n} T^2 \\ &=& \delta(X_1,\ldots,X_n) \end{eqnarray} $$ Here $X_{k:n}$ denotes $k$-th out of $n$ order statistics.

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  • $\begingroup$ How can I use the the relation with the Cramer-Rao Lower Bond condition? That is if $\hatT$ is unbaised for $\T(\theta)$ and $var(\hatT)$ = CRLB, then $\hatT$ is the UMVUE of $T(\theta). $\endgroup$ – David Feb 11 '12 at 21:04

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