Statistik 1

Aufgabe 1
\(\mathbb{E}_\theta[\sum_{i=1}^{n}a_ix_i]=\sum_{i=1}^na_i\mathbb{E}_{\theta}[X_i]=\mathbb{E}_\theta[X_1]*\sum_{i=1}^na_i\\ \\ => \mathbb{E}_\theta[\sum_{i=1}^{n}a_ix_i]=\mathbb{E}_\theta[X_1]\ <=>\ \sum_{i=1}^na_i=1\)




\(a_1=\frac{1}{m}+x,\ a_2=\frac{1}{m}-x,\ a_i=c\)




\(M:=\{i\in \mathbb{N} | a_i \neq 0 \},\ |M| = m \leq n\)
\(\forall i\in M: a_i=\frac{1}{m}:\)


\(\mathbb{V}\delta=\sum_{i=1}^{n}a_i^2\mathbb{V}X_i + \sum_{i\neq j}a_ia_j\mathbb{V}X_i\)

\(=\mathbb{V}X_1(\sum_{i=1}^{m}\frac{1}{m^2} + \sum_{i\neq j}\frac{1}{m^2})\)

\(=\mathbb{V}X_1(\frac{1}{m} + \frac{m!}{(m-2)!m^2})\)

\(=\mathbb{V}X_1(\frac{1}{m} + \frac{m-1}{m})\)

\(=\mathbb{V}X_1\)



\(\mathbb{V}\delta=\mathbb{V}X_1(\sum_{i=1}^{n}a_i^2 + \sum_{i\neq j}a_ia_j)\)

\((\sum_{i=1}^{n}a_i^2 + \sum_{i\neq j}a_ia_j)\neq 1\)



IA: \(m=2\)

\(c = \sum_{i=1}^{m}\)

\(\sum_{i=1}^{2}a_i^2 + \sum_{i\neq j}a_ia_j = c^2\)

\(\Leftrightarrow a_1^2+a_2^2+2a_1a_2 = c^2\)

\(\Leftrightarrow a_1^2+(c-a_1)^2+2a_1(c-a_1) = c^2\)

\(\Leftrightarrow a_1^2+c^2-2a_1+a_1^2+2a_1-2a_1^2=c^2\)

Aufgabe 2
\(p_i(X_1,..,X_n) = X_i*(1-e^{-\theta})\)

\(\mathbb{E}_\theta[p_i]=\theta\)

\(\Leftrightarrow\sum_{n\in\mathbb{N}}n*(1-e^{-\theta})*\frac{\theta^n}{n!}*e^{-\theta}*\frac{1}{1-e^{-\theta}}=\theta\)

\(\Leftrightarrow e^{-\theta}*\frac{1}{1-e^{-\theta}}*\sum_{n\in\mathbb{N}}n*(1-e^{-\theta})*\frac{\theta^n}{n!}=\theta\)

\(\Leftrightarrow e^{-\theta}*\sum_{n\in\mathbb{N}}n*\frac{\theta^n}{n!}=\theta\)

\(\Leftrightarrow e^{-\theta}*\theta*\sum_{i=1}^{\infty}\frac{\theta^{n-1}}{(n-1)!}=\theta\)

\(\Leftrightarrow e^{-\theta}*\theta*\sum_{i=0}^{\infty}\frac{\theta^n}{n!}=\theta\)

\(\Leftrightarrow e^{-\theta}*\theta*e^\theta=\theta\)

\(\Leftrightarrow \theta=\theta\)