Mathematische Statistik - Übung 6.39

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Seien X_1, X_2, \ldots, X_n \, unabhängig und Poisson-verteilt mit Parameter \lambda \,. Zeigen sie

T(X) = \frac 1 n \sum_{i=1}^n x_i \,

ist suffiziente Statistik für \lambda \, (insbesonders minimal suffiziente Statistik) 

 T \, suffizient  \Leftrightarrow P_{\theta}(x) = g(T(x), \theta) h(x) \,

 
\begin{align}
P_\theta(X)
& = \prod_{i=1}^n \frac{\theta^{x_i}}{x_i!} e^{-\theta} \\
& = \underbrace{e^{-n \theta} \theta^{n \frac 1 n \sum_{i=1}^n x_i}}_{g(T(x), \theta)} \underbrace{\frac{1}{\prod_{i=1}^n x_i!}}_{h(x)}
\end{align}
\,

minimal suffiziente Statistik

 
\begin{align}
e^{-n \theta} \theta^{n T} \frac{1}{\prod_{i=1}^n x_i!} & = & g(S(x), \theta) h(x) \\
\end{align}
\,

 
\begin{array}{rcl}
\frac{e^{-n \theta_1} \theta_1^{n T} \frac{1}{\prod_{i=1}^n x_i!}}
     {e^{-n \theta_2} \theta_2^{n T} \frac{1}{\prod_{i=1}^n x_i!}}
 & = & 
\frac{g(S(x), \theta_1) h(x)}
     {g(S(x), \theta_2) h(x)}
\\
e^{-n \theta_1 + n \theta_2} \left(\frac{\theta_1}{\theta_2}\right)^{n T}
 & = & 
\frac{g(S(x), \theta_1)}
     {g(S(x), \theta_2)}
\\
T n \log  \left(\frac{\theta_1}{\theta_2}\right) -n \theta_1 + n \theta_2
 & = & 
\log \left(
\frac{g(S(x), \theta_1)}
     {g(S(x), \theta_2)}
\right)
\\
T & = & 
\frac{
	\log \left(
	\frac{g(S(x), \theta_1)}
	     {g(S(x), \theta_2)}
	\right)
	+ n \theta_1 - n \theta_2
}
{
n \log  \left(\frac{\theta_1}{\theta_2}\right)
}
\end{array}
 \,
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