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The unbiasedness part holds trivially from the fact that you have stipulated that the unknown quantity you are estimating is the expected value of the estimator. Denoting this by the parameter $\theta \equiv \mathbb{E}_\varphi(T(\mathbf{X}))$ and noting that your estimator is $\hat{\theta} \equiv T(\mathbf{X})$, you then clearly have: \text{Bias}(\hat{\... 3 Randomised estimators are defined in a more general setting than in this case and in particular need not be connected with sufficiency. In full generality, a randomised decision rule is a decision rule that returns a random decision for a given observation (or dataset). To reproduce the quote from Lehmann and Casella : The starting point of a statistical ... 0 Suppose X_1,X_2,\ldots,X_n are i.i.d N(\theta,1) where \theta\in \mathbb R. Then \overline X is a complete sufficient statistic for \theta. Note that X_{(1)}-\overline X=(X_{(1)}-\theta)-(\overline X-\theta), so that its distribution is free of \theta. By Basu's theorem, the ancillary statistic X_{(1)}-\overline X is independent of \overline ... 4 No. You can multiply a sufficient estimator by anything (like, say, 𝑛/(𝑛−1)) and it will stay sufficient but stop being unbiased 0 Pollard argues that disintegrations, which give conditional probability distributions under fairly general conditions, are the right way to think about the factorisation theorem and sufficiency in the continuous case. 0 An estimator \hat{\theta} of \theta may (correctly) depend on features of the data that provide information on \theta (i.e. depend on the sufficient statistic T(X)), but it may also, in general, depend on features that do not provide information on \theta. The latter dependence is a source of variance that we'd like to eliminate -- there is no ... 0 Here is my attempt. 1.Show that f(x; \varphi) is in the one-parameter exponential family. Assume that \varphi > 0. Rewriting f(x; \varphi), we have that \begin{align*} f(x; \varphi) &= \exp \left\{\log \left[ \varphi x^{\varphi - 1} \cdot \mathbb{I}(0 < x < 1) \right] \right\} \\ &= \exp \left\{ \log \varphi + (\varphi - 1) \log x + \... 0 By writing the joint density in exponential family form we find f(x,\theta) = \frac1{\sqrt{2\pi}} \exp\left\{ -\frac{n}{2}\log\theta - \frac1{2\theta} \sum_i x_i^2 + \sum_i x_i -n\theta^2 \right\} $$and then the result follows by general exponential family theory, as noted in comments. 2 To apply¹ Theorem 9.36, the relevant implication for minimality is that x^n \leftrightarrow y^n implies T(x^n) = T(y^n) (as the reverse implication always holds for any sufficient statistic T(\mathbf y)). Hence if one starts from the property x^n \leftrightarrow y^n holding for a given pair (\mathbf y,\mathbf x), it means that$$L(\mu;\mathbf y)=c(...

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In AoS, there are definitions and there are theorems. Theorems are helpful in identifying properties. When achieving the equality $$L(\mu;\mathbf y) = (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}}$$ the exponential can be broken into parts \begin{align} L(\mu;\mathbf y) =&...

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In that book, $n$ random variables $X_1, \dots, X_n$, constituting the "data set", are represented as $X^n$. Whereas realisations of those $n$ random variables $x_1, x_2, \dots , x_n$ are represented as $x^n$. There is no exponentiation occurring.

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