关于无偏估计量定义的理解及样本均值无偏性证明的困惑
Hey there, let's work through your confusion about unbiased estimators step by step. First, let's lay out your current understanding and the sticking point clearly:
背景定义回顾
First, here's the setup I'm working with:
- We have a random sample $X_1,...,X_n$ of real-valued random variables defined on a common probability space.
- A statistic is a measurable function $g:\mathbb{R}^n\to\mathbb{R}$ applied to the sample, written as $g(X_1,...,X_n)$. When this is used to estimate a parameter $\theta$ of the common distribution of the $X_i$, we call it a point estimator—often denoted $\hat{\Theta}(X_1,...,X_n)$ or $g(X_1,...,X_n;\theta)$.
The formal definition of an unbiased estimator is:
Let $\varphi:\text{parameters}\to\text{probability measures on } \mathbb{R}^n$ be a parameterization where $\varphi(\theta)=\mu_{\theta}$. For an estimator $\hat{\Theta}=g(X_1,...,X_n)$ of $\theta$, we define its expectation under $\theta$ as:
$$E_{\theta}(\hat{\Theta})=\int_{\mathbb{R}^n}g(x_1,...,x_n)d\mu_{\theta}(x_1,...,x_n)$$
$\hat{\Theta}$ is unbiased if $E_{\theta}(\hat{\Theta})=\theta$ for every $\theta$.
我的问题：样本均值的无偏性证明
I need to show that if $\theta=E(X_1)$, then the sample mean $\hat{\Theta}=\frac{1}{n}\sum\limits_{i=1}^nX_i$ is an unbiased estimator of $\theta$.
I get the general idea—using linearity of the integral and the fact that the $X_i$ are identically distributed—but I'm confused about how to handle the measure $\mu_{\theta}$ here. I've started the proof like this:
$$
\begin{align*}
E_{\theta}(\hat{\Theta})&=\int_{\mathbb{R}^n}g(x_1,...,x_n)d\mu_{\theta}(x_1,...,x_n)\
&=\frac{1}{n}\int_{\mathbb{R}n}\sum_{i=1}nx_id\mu_{\theta}(x_1,...,x_n)
\end{align*}
$$
I want to split this into a sum of $n$ integrals, each equal to $\theta$, but I don't see how to justify that with the product measure $\mu_{\theta}$. Am I on the right track so far, or am I misunderstanding the definition of $E_{\theta}$?
专家解答
Great question—you're totally on the right track! The key missing piece here is remembering that for a standard random sample (which implies independence), the joint measure $\mu_{\theta}$ on $\mathbb{R}^n$ is the product measure of the marginal measure $\nu_{\theta}$ (where $\nu_{\theta}$ is the distribution of each individual $X_i$, so $\theta = \int_{\mathbb{R}} x d\nu_{\theta}(x)$).
Let's break down the proof step by step to clear up the measure confusion:
Swap sum and integral: First, we can use linearity of integration to move the sum outside the integral—this is valid even for multiple integrals:
$$
\frac{1}{n}\int_{\mathbb{R}n}\sum_{i=1}nx_id\mu_{\theta}(x_1,...,x_n) = \frac{1}{n}\sum_{i=1}n\int_{\mathbb{R}n}x_id\mu_{\theta}(x_1,...,x_n)
$$Leverage product measure with Fubini's theorem: Take any single term in the sum, say the $k$-th integral: $\int_{\mathbb{R}^n}x_k d\mu_{\theta}(x_1,...,x_n)$. Since $\mu_{\theta} = \nu_{\theta} \times \nu_{\theta} \times ... \times \nu_{\theta}$ (the product of $n$ copies of the marginal measure), we can apply Fubini's theorem (valid here because $x_k$ is integrable—we know $\theta = E(X_k)$ exists) to rewrite it as an iterated integral:
$$
\int_{\mathbb{R}^n}x_k d\mu_{\theta}(x_1,...,x_n) = \int_{\mathbb{R}}...\int_{\mathbb{R}}x_k d\nu_{\theta}(x_1)d\nu_{\theta}(x_2)...d\nu_{\theta}(x_n)
$$Simplify the iterated integral: Notice that $x_k$ doesn't depend on $x_1,...,x_{k-1},x_{k+1},...,x_n$. We can split this into a product of separate integrals:
$$
\left(\int_{\mathbb{R}}d\nu_{\theta}(x_1)\right)...\left(\int_{\mathbb{R}}x_k d\nu_{\theta}(x_k)\right)...\left(\int_{\mathbb{R}}d\nu_{\theta}(x_n)\right)
$$
Each integral over $x_i$ (where $i \neq k$) is just the total probability of the marginal distribution, which equals 1. The integral over $x_k$ is exactly $\theta = E(X_k)$. So every term in our sum equals $\theta$.Final calculation: We have $n$ terms each equal to $\theta$, so:
$$
\frac{1}{n}\sum_{i=1}^n\theta = \frac{1}{n} \times n\theta = \theta
$$
This proves $E_{\theta}(\hat{\Theta}) = \theta$, so the sample mean is indeed an unbiased estimator.
To wrap up: Your initial step was perfect—you just needed to connect the abstract joint measure $\mu_{\theta}$ to the product of the marginal measures of each $X_i$. Once you make that link, linearity of integration and Fubini's theorem do the rest to simplify the integral into something intuitive.
备注：内容来源于stack exchange，提问作者user124910
