Neyman Theorem

Theorem

Let

• $X_1,\dots ,X_n=\mathbf{X}$ : Random sample, with distribution that has pdf/pmf $f(x;\theta )$, $\theta \in \Omega$
• $Y=u(\mathbf{X})$ : Statistic for $\theta$
• $L(\theta )$ : Likelihood function of $X$

Then $Y$ is a sufficient statistic for $\theta$

If and only if

• $\exists k,l\ni L(\theta )=k[u(\mathbf{x});\theta ]\cdot l(\mathbf{x})$
• $l(\mathbf{x})$ is independent of $\theta$

Remark

To determine if $Y$ is a sufficient statistic,

we have to show that its likelihood function

can be written as the multiplication of two functions:

1. A function of $Y$ that may depend on $\theta$
2. A function of the random sample that DOES NOT depend on $\theta$

Example

Let $X_1,X_2,\dots ,X_n$ represent a random sample from the Poisson distribution with parameter $\lambda$. Prove that both $Y_1=\sum X_i$ and $Y_2=\bar{x}$ are sufficient statistics for $\lambda$ using the factorization theorem.

$$\begin{array}{rl}L(\theta )& =\prod _{i=1}^n\frac{e^{-\theta }{\theta }^{x_i}}{x_i!}\\ & =e^{-n\theta }{\theta }^{{\sum }_{i=1}^n{x}_i}\cdot \prod _{i=1}^n{x}_i!\end{array}$$

Let

• $k_1[u(X);\theta ]=e^{-n\theta }{\theta }^{{\sum }_{i=1}^n{x}_i}$
• $k_2(x)={\prod }_{i=1}^n{x}_i!$

Notice that $u(X)$ is a function of $Y_1=\sum X_i$. Also, because $Y_2=\frac{Y_1}{n}$, then $u(X)$ is also a function of $u(X)$ (or that $k_1:e^{-n\theta }{\theta }^{n\bar{x}}$).