Retaining the estimator exactly as printed, define the probability limit
$$
c=\frac{\mathbb E[ZX]}{\mathbb E[Z^2]},
\qquad W=Z-cX.
$$
The empirical equations are linear in $(\beta,\alpha)$. Their coefficient matrix converges to
$$
M=
\begin{pmatrix}
\mathbb E[AW]&\mathbb E[XW]\\
\mathbb E[AX]&\mathbb E[X^2]
\end{pmatrix}.
$$
Therefore a sufficient condition, requiring neither parametric assumption, is
$$
0<\mathbb E[Z^2]<\infty,
\qquad
\det M
=\mathbb E[AW]\mathbb E[X^2]
-\mathbb E[XW]\mathbb E[AX]\ne0,
$$
with finite moments sufficient for the <weak law of large numbers>. The empirical determinant then converges to a nonzero number, so the two linear equations have a unique solution with probability tending to one.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 221 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-221/2/b/solution