1.
Pour y > 0 y\gt 0 y > 0 Y = X 3 θ Y=\dfrac{X^3}{\theta} Y = θ X 3 x = ( θ y ) 1 / 3 x=(\theta y)^{1/3} x = ( θ y ) 1/3 d x = 1 3 θ 1 / 3 y − 2 / 3 d y dx=\tfrac{1}{3}\theta^{1/3}y^{-2/3}dy d x = 3 1 θ 1/3 y − 2/3 d y
f Y ( y ) = 3 ( θ y ) 2 / 3 θ e − y ⋅ 1 3 θ 1 / 3 y − 2 / 3 = e − y . f_Y(y)=\frac{3(\theta y)^{2/3}}{\theta}e^{-y}\cdot\tfrac{1}{3}\theta^{1/3}y^{-2/3}=e^{-y}. f Y ( y ) = θ 3 ( θ y ) 2/3 e − y ⋅ 3 1 θ 1/3 y − 2/3 = e − y .
Donc Y ∼ E ( 1 ) Y\sim\mathcal{E}(1) Y ∼ E ( 1 ) E ( Y ) = 1 \mathbb{E}(Y)=1 E ( Y ) = 1 Var ( Y ) = 1 \operatorname{Var}(Y)=1 Var ( Y ) = 1
Comme X 3 = θ Y X^3=\theta Y X 3 = θ Y E ( X 3 ) = θ \mathbb{E}(X^3)=\theta E ( X 3 ) = θ Var ( X 3 ) = θ 2 \operatorname{Var}(X^3)=\theta^2 Var ( X 3 ) = θ 2
2.
Vraisemblance : L ( θ ) = ∏ i = 1 n 3 x i 2 θ e − x i 3 / θ L(\theta)=\prod_{i=1}^n \dfrac{3x_i^2}{\theta}e^{-x_i^3/\theta} L ( θ ) = ∏ i = 1 n θ 3 x i 2 e − x i 3 / θ
ℓ ( θ ) = cte − n ln θ − 1 θ ∑ i = 1 n x i 3 . \ell(\theta)=\text{cte}-n\ln\theta-\frac{1}{\theta}\sum_{i=1}^n x_i^3. ℓ ( θ ) = cte − n ln θ − θ 1 ∑ i = 1 n x i 3 .
d ℓ d θ = − n θ + 1 θ 2 ∑ x i 3 = 0 \dfrac{d\ell}{d\theta}=-\dfrac{n}{\theta}+\dfrac{1}{\theta^2}\sum x_i^3=0 d θ d ℓ = − θ n + θ 2 1 ∑ x i 3 = 0
θ ^ n = 1 n ∑ i = 1 n X i 3 . \boxed{\widehat{\theta}_n=\frac{1}{n}\sum_{i=1}^n X_i^3.} θ n = n 1 i = 1 ∑ n X i 3 .
3.
E ( θ ^ n ) = 1 n ∑ E ( X i 3 ) = θ \mathbb{E}(\widehat{\theta}_n)=\dfrac{1}{n}\sum \mathbb{E}(X_i^3)=\theta E ( θ n ) = n 1 ∑ E ( X i 3 ) = θ sans biais .
Var ( θ ^ n ) = 1 n 2 ∑ Var ( X i 3 ) = θ 2 n → 0 \operatorname{Var}(\widehat{\theta}_n)=\dfrac{1}{n^2}\sum\operatorname{Var}(X_i^3)=\dfrac{\theta^2}{n}\to 0 Var ( θ n ) = n 2 1 ∑ Var ( X i 3 ) = n θ 2 → 0 θ ^ n \widehat{\theta}_n θ n convergent (sans biais et variance tendant vers 0).