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I have some confusion with the definition of unbiased estimators. Suppose you are given a random sample $X_1,...,X_n$ of real-valued random variables defined on a common probability space. A statistic is defined as $g(X_1,...,X_n)$, where $g:\mathbb{R}^n\to\mathbb{R}$ is a measurable function. If there is some parameter $\theta$ of the common distribution of $X_1,...,X_n$ that we want to approximate, we might emphasize this with the notation $g(X_1,...,X_n;\theta)$, or $\hat{\Theta}(X_1,...,X_n)$, in which case, we call this random variable a point-estimator.

Definition: Let $\varphi:\text{parameters}\to\text{probility measures on} \ \mathbb{R}^n$, be a parameterization $\varphi(\theta)=\mu_{\theta}$. If $\hat{\Theta}=g(X_1,...,X_n)$ is an estimator of $\theta$, define $E_{\theta}(\hat{\Theta})=\int_{\mathbb{R}^n}g(x_1,...,x_n)d\mu_{\theta}(x_1,...,x_n)$. $\hat{\Theta}$ is said to be an unbiased estimator if $E_{\theta}(\hat{\Theta})=\theta$, for each $\theta$.

Question and example: Show that if $\theta=E(X_1)$, then $\hat{\Theta}=\frac{1}{n}\sum\limits_{i=1}^nX_i$ is an unbiased estimator of $\theta$.

I understand the idea is just using the linearity of the integral and the fact that $X_1,...,X_n$ are identically distributed, but I'm a little confused by these different measures. $\theta$ is mapped to some probability measure $\mu_{\theta}$ on $\mathbb{R}^n$. We must show with respect to this measure, $E_{\theta}(\hat{\Theta})=\theta$. \begin{align*} E_{\theta}(\hat{\Theta})&=\int_{\mathbb{R}^n}g(x_1,...,x_n)d\mu_{\theta}(x_1,...,x_n)\\ &=\frac{1}{n}\int_{\mathbb{R}^n}\sum_{i=1}^nx_id\mu_{\theta}(x_1,...,x_n) \end{align*}$

I'd like to write this as a sum of $n$ integrals, each with integral $\theta$, but I'm not seeing how to do this with this product measure. Is this correct so far, or am I not understanding the definition of $E_{\theta}$ correctly?

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Given a statistical model $(\Omega, P_{\theta})$ parameterized by $\theta\in \Theta\subset R^d$ an estimator of $\theta$ is $any$ function $g$ from $\Omega$ to $R^d$, $w\mapsto g(w).$ It is said to be unbiased when $$E(g(w))=\int_{\Omega}g(w)P_{\theta}(dw)$$ does exist and is equal to $\theta$ for all $\theta\in \Theta.$ The function $g$ does not depend on $\theta$, this is crucial.

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  • $\begingroup$ It's probably not great notation I used: $g(X_1,..., X_n;\theta)$. My understanding is, when you're emphasizing you want to approximate a parameter, that makes a statistic an estimator. Not that the function itself depends on $\theta$. $\endgroup$
    – user124910
    Commented Feb 20 at 23:55
  • $\begingroup$ In this particular example, $\Omega$ is $\mathbb{R}^d$ and $\mathbb{R}^d$ is just $\mathbb{R}$, for the sample to mean as an estimator for $E(X_1)$, correct? I'm not seeing how $\int_{\Omega}g(\omega)P_{\theta}(d\omega)=E(X_1)$, if $\theta=E(X_1)$. $P_{\theta}$ could be any measure on $\mathbb{R}^n$, depending on the parameterization. $\endgroup$
    – user124910
    Commented Feb 21 at 0:00

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