Lehmann-Scheffé Theorem

1. Statement

Theorem (Lehmann-Scheffé Theorem). Let $\mathbf{X}=(X_1,\cdots ,X_n{)}^T$ be a random sample from a distribution with pdf $f(\mathbf{x}|\theta )$. Suppose $T(\mathbf{X})$ is a complete and sufficient statistic for $\theta$, and $h(T(\mathbf{X}))$ is unbiased for $\theta$. Then $h(T(\mathbf{X}))$ is the unique UMVUE of $\theta$.

2. Proof

Proof

Given such $T$, we have to show that for all $\hat{\theta }$ such that $\mathbb{E}[\hat{\theta }]=\theta$, we have $\text{MSE}(h(T))\le \text{MSE}(\hat{\theta })$.

According to the Rao-Blackwell Theorem,

$${\hat{\theta }}_{RB}=\mathbb{E}[\hat{\theta }|T]=\tilde{\theta }(T).$$

Then it is a better estimator than $\hat{\theta }$:

$$\text{MSE}({\hat{\theta }}_{RB})=\text{MSE}(\tilde{\theta }(T))\le \text{MSE}(\hat{\theta }).$$

Moreover, both ${\hat{\theta }}_{RB}$ and $h(T)$ are unbiased for $\theta$:

$$\mathbb{E}[h(T)-\tilde{\theta }(T)]=0.$$

Since $T$ is complete and $h(T)-\tilde{\theta }(T)$ is a function of $T$, we have $h(T)=\tilde{\theta }(T)$ almost everywhere.

Therefore, $h(T)$ is the unique UMVUE of $\theta$:

$$\text{MSE}(h(T))=\text{MSE}(\tilde{\theta }(T))\le \text{MSE}(\hat{\theta }).$$

$\square$

Related Notes

• Rao-Blackwell Theorem
• Sufficiency
• UMVUE