Loi normale bivariée

Préliminaire 1

Maths stack exchange

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.

1a) $EY=a$ donc $\hat{a_Y} = \overline{Y}$

1b) $Var(\hat{a_Y}) = \frac{ b \mu ^2 +\sigma^2}{n}$

2a) $E(XY)= b E(X^2)$ donc $\hat{b} = \dfrac{ \sum X_i^2}{n}$

3a)
$\hat{b_0} = \dfrac{ b \left( \tfrac{ \sum{X_i^2} }{n} - \Big(\tfrac{\sum X_i}{n} \Big)^2 \right)
+ b \left( \tfrac{ \sum {X_iY_i} }{n} - \tfrac{\sum X_i}{n}
\tfrac{\sum U_i}{n} \right) } { \tfrac{ \sum \Big( X_i - \tfrac{ \sum X_i }{n} \Big) ^2 }{n} }$

En fait, on a $ \hat{b}_0= b + \dfrac{ \sum (X_i - \bar{X} ) U_i }{ \sum ( X_i - \bar{X})^2} := b +Z $
$V(Z)= E(V(Z|X)) + V( E(Z|X) ) =E ( V (Z|X)) = \dfrac{\sigma^2}{ \mu^2} $

5 a) $ \mathcal{N} (x_i, \beta^2)$

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.