maths stack exchange a écrit: $$var(X_1 X_2) = E(var(X_1 X_2|X_2))+var(E(X_1 X_2 | X_2)\\ =E(X_2^2 var(X_1)) + var(X_2 E(X_1))\\ = var(X_1) (E(X_2)^2 + var(X_2)) + E(X_1)^2 var(X_2)\\ = var(X_1) var(X_2) + E(X_2)^2 var(X_1) + E(X_1)^2 var(X_2)$$ Since $X_1, X_2$ are not constants then we must have $E(X_1)=E(X_2)=0$ which is necessary and sufficient.
Réponses
Maths stack exchange
$(X,Y)^t = A (X,U)^t + (0,a) $ avec $A=\begin{pmatrix} \mu^2& 0 \\ 0 & \sigma^2 \end{pmatrix}$
donc $Var \{ (X,Y)^t \} = A Var( (X,U)^t ) A' = \begin{pmatrix} \mu^2& b \mu^2 \\ b \mu^2 & b \mu^2 + \sigma^2 \end{pmatrix} : = \Sigma$
1b) $Var(\hat{a_Y}) = \frac{ b \mu ^2 +\sigma^2}{n}$
$\hat{b_0} = \dfrac{ b \left( \tfrac{ \sum{X_i^2} }{n} - \Big(\tfrac{\sum X_i}{n} \Big)^2 \right)
+ b \left( \tfrac{ \sum {X_iY_i} }{n} - \tfrac{\sum X_i}{n}
\tfrac{\sum U_i}{n} \right) } { \tfrac{ \sum \Big( X_i - \tfrac{ \sum X_i }{n} \Big) ^2 }{n} }$
$V(Z)= E(V(Z|X)) + V( E(Z|X) ) =E ( V (Z|X)) = \dfrac{\sigma^2}{ \mu^2} $
a + \dfrac { \sum X_i}{n} Z = \dfrac{1}{ \mu^2} ( \overline{XU} - \overline{X}^2 \overline{U} )$
5 b) $U_i^{\star} = b ( X_i-X_i ^\star) + U_i \sim \mathcal {N}( 0 , b^2 \beta^2 +\sigma ^2)$
5 c) $cov(U_i ^ \star , X_i ^\star) \ne 0$ donc l'estimateur dans ce modèle n'est pas convergent.