1

$Ω = {(x_{1}, ... x_{10}) : x \in {n, m, h}}$
$∣Ω∣ = 3^{10}$

∣ A ∣ = (3 10) (3 7) = \frac{10 !}{( 10 - 3 )! 3 !} \frac{7 !}{( 7 - 3 )! 3 !} = \frac{10 ! \cdot 7 !}{7 ! \cdot 3 ! \cdot 4 ! \cdot 3 !} = \frac{10 !}{2 \cdot 3 ! \cdot 4 !}

P (A) = \frac{∣ A ∣}{∣Ω∣} = \frac{( 3 10 ) ( 3 7 )}{3 ^{10}} = \frac{10 !}{2 \cdot 3 ! \cdot 4 !} \cdot \frac{1}{3 ^{10}} = \frac{10 !}{2 \cdot 3 ! \cdot 4 ! \cdot 3 ^{10}}

2

\int_{Ω} f_{X (x)} d x \int_{0}^{1} γ d x γ \cdot (1 - 0) γ = 1 = 1 = 1 = 1

E [X] = \int_{- \infty}^{\infty} x \cdot f_{X} (x) d x = \int_{0}^{1} x \cdot γ d x = \int_{0}^{1} x d x = [\frac{1}{2} x^{2}]_{0}^{1} = \frac{1}{2} 1^{2} - \frac{1}{2} 0^{2} = \frac{1}{2} 1 - \frac{1}{2} 0 = \frac{1}{2}

E [X^{2}] = \int_{- \infty}^{\infty} x^{2} \cdot f_{X} (x) d x = \int_{0}^{1} x^{2} d x = [\frac{1}{3} x^{3}]_{0}^{1} = \frac{1}{3}

Va r (X) = E [X^{2}] - (E [X])^{2} = \frac{1}{3} - (\frac{1}{2})^{2} = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}

\overset{p}{^}_{1} = \frac{N _{1}}{n} \overset{p}{^}_{2} = \frac{N _{2}}{n} \overset{p}{^}_{3} = \frac{N _{3}}{n}

p_{1} (θ) + p_{2} (θ) + p_{3} (θ) = 1

gilt

p_{1} (θ) = 1 - p_{2} (θ) - p_{3} (θ)

da $p_{1} (θ) = θ$ ist gilt

θ = 1 - p_{2} (θ) - p_{3} (θ)

und so für den Schätzer

\hat{θ} = 1 - \overset{p}{^}_{2} (θ) - \overset{p}{^}_{3} (θ) = 1 - \frac{N _{2}}{n} - \frac{N _{3}}{n}

y = {y_{1}, ..., y_{n}}^{T}

Z (y, θ) = θ^{n_{1}} \cdot (α (1 - θ))^{n_{2}} \cdot ((1 - α) \cdot (1 - θ))^{n_{3}}

$n_{j} = ∣ {1 \leq i \leq n : y_{i} = j} ∣$
$j \in {1, 2, 3}$

L (y, θ) = l o g (p_{1} (y_{1}, θ)) + l o g (p_{2} (y_{2}, θ)) + l o g (p_{3} (y_{3}, θ)) = n_{1} l o g (θ) + n_{2} l o g (α) + n_{2} l o g (1 - θ) + n_{3} l o g (1 - α) + n_{3} l o g (1 - θ)

Um das Maximum zu finden müssen wir nach $θ$ ableiten.

\frac{d L ( y , θ )}{d θ} = n_{1} \frac{1}{θ} + n_{2} \frac{- 1}{1 - θ} + n_{3} \frac{- 1}{1 - θ} = n_{1} \frac{1}{θ} + (n_{2} + n_{3}) \frac{- 1}{1 - θ} = \frac{n _{1}}{θ} - \frac{n _{2} + n _{3}}{1 - θ}

\frac{d L ( y , θ )}{d θ} = 0 ⟺ \frac{n _{1}}{θ} = \frac{n _{2} + n _{3}}{1 - θ} ⟺ \frac{θ ( 1 - θ ) n _{1}}{θ} = \frac{θ ( 1 - θ ) ( n _{2} + n _{3} )}{1 - θ} ⟺ (1 - θ) n_{1} = θ (n_{2} + n_{3}) ⟺ n_{1} - θ n_{1} = θ n_{2} + θ n_{3} ⟺ n_{1} = θ n_{1} + θ n_{2} + θ n_{3} ⟺ n_{1} = θ (n_{1} + n_{2} + n_{3}) ⟺ n_{1} = θ n ⟺ θ = \frac{n _{1}}{n}

0^{k} = 0 \land 1^{k} = 1

Nein da die Dichte Funktion auch unbeschränkt sein kann.
$Va r (X) = E [X^{2}] - E^{2} [X]$

f_{X (x)} = c

E (X^{2}) = \int_{a}^{b} x^{2} \cdot c d x

E^{2} (X) = (\int_{a}^{b} x \cdot c d x)^{2} = \int_{a}^{b} x^{2} \cdot c^{2} d x

E (X^{2}) \leq E^{2} (X)