证明正态分布样本统计量T=($\sum_{i}X_{i},\sum_i X_{i}^{2}$)在n≥2时不完备

Alright, let's walk through this proof step by step, starting with the core definition we need to work with.

First, let's recap the definition of a complete statistic (this is the key to our argument):

A statistic $T$ is complete if for every measurable function $g$, the condition $\mathbb{E}{\theta}[g(T)] = 0$ for all $\theta$ in the parameter space implies that $P{\theta}(g(T) = 0) = 1$ for all $\theta$.

In plain terms: to show $T$ is not complete, we just need to find a non-trivial measurable function $g(T)$ (one that isn't zero almost everywhere) such that its expectation is zero for every possible $\theta$.

Step 1: Construct the function $g(T)$ using the given hint #

The problem gives us a critical clue: we know $\mathbb{E}{\theta}\left[2\left(\sum X_i\right)^2 - (n+1)\sum X_i^2\right] = 0$ for all $\theta$. Let's define our function directly from this expression:
$$g(T) = 2\left(\sum{i=1}^n X_i\right)^2 - (n+1)\sum_{i=1}^n X_i^2$$
Since $T = \left(\sum X_i, \sum X_i^2\right)$, $g(T)$ clearly only depends on our statistic $T$, so it's a valid function to use for our proof.

Step 2: Verify $\mathbb{E}_{\theta}[g(T)] = 0$ for all $\theta$ #

We can confirm this expectation is zero using basic properties of the normal distribution:

• For each $X_i \sim N(\theta, \theta^2)$, we have $\mathbb{E}[X_i] = \theta$, and $\mathbb{E}[X_i^2] = \text{Var}(X_i) + (\mathbb{E}[X_i])^2 = \theta^2 + \theta^2 = 2\theta^2$.
• For the sum $\sum X_i$, its expectation is $n\theta$, and its variance is $n\theta^2$, so $\mathbb{E}\left[\left(\sum X_i\right)^2\right] = \text{Var}\left(\sum X_i\right) + \left(\mathbb{E}\sum X_i\right)^2 = n\theta^2 + (n\theta)^2 = n(n+1)\theta^2$.

Calculating the expectation of $g(T)$:

\begin{align*}
\mathbb{E}[g(T)] &= 2\mathbb{E}\left[\left(\sum X_i\right)^2\right] - (n+1)\mathbb{E}\left[\sum X_i^2\right] \\
&= 2 \cdot n(n+1)\theta^2 - (n+1) \cdot n \cdot 2\theta^2 \\
&= 2n(n+1)\theta^2 - 2n(n+1)\theta^2 \\
&= 0
\end{align*}

This holds for every $\theta$, so we've satisfied the first condition needed to disprove completeness.

Step 3: Show $g(T)$ is not zero almost everywhere #

Now we need to prove that $g(T)$ isn't trivially zero (i.e., there exists some $\theta$ where $P_{\theta}(g(T) \neq 0) > 0$):
Take $\theta \neq 0$, and consider the case where all sample values equal $\theta$ (this has positive probability because the normal distribution has positive density at $\theta$). For this sample:

• $\sum X_i = n\theta$
• $\sum X_i^2 = n\theta^2$

Plugging these into $g(T)$:
$$g(T) = 2(n\theta)^2 - (n+1)n\theta^2 = 2n^2\theta2 - n(n+1)\theta^2 = n\theta^2(n-1)$$
Since $n \ge 2$, $n-1 \ge 1$, so this value is non-zero (as long as $\theta \neq 0$). This means there's a positive-probability event where $g(T) \neq 0$, so $g(T)$ is not zero almost everywhere.

Conclusion #

We've found a non-trivial function $g(T)$ such that $\mathbb{E}_{\theta}[g(T)] = 0$ for all $\theta$, but $g(T)$ isn't zero almost everywhere. By the definition of completeness, this means $T = \left(\sum X_i, \sum X_i^2\right)$ is not a complete statistic when $n \ge 2$.

内容的提问来源于stack exchange，提问作者Galois