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\textbf{Aufgabe 1}\\  $\mathbb{E}_\theta[\sum_{i=1}^{n}a_ix_i]=\sum_{i=1}^na_i\mathbb{E}_{\theta}[X_i]=\mathbb{E}_\theta[X_1]*\sum_{i=1}^na_i\\  \\  => \mathbb{E}_\theta[\sum_{i=1}^{n}a_ix_i]=\mathbb{E}_\theta[X_1]\ <=>\ \sum_{i=1}^na_i=1$ \sum_{i=1}^na_i=1$\\  \\  \\  \\  $a_i=1/n$:\\  $\sum_{i=1}^{n}\frac{1}{n^2}*\mathbb{V}X_i + \sum_{i\neq j}\frac{1}{n^2}*\mathbb{V}X_i$