Ignoring constants, the complete-data log posterior is
$$
-\frac1{2\sigma^2}\|Y-X\beta\|^2
-\frac1{2\sigma_\beta^2}\|\beta\|^2
+\sum_{i,j}\{X_{ij}\log\pi_j+(1-X_{ij})\log(1-\pi_j)\}.
$$
Take conditional expectations under $(\beta^{(t)},\pi^{(t)})$. Define
$$
A=\frac1{2\sigma^2}\mathbb E[X^TX\mid Y,X_O,\theta^{(t)}]
+\frac1{2\sigma_\beta^2}I,
$$
$$
d=A^{-1}\frac{\mathbb E[X\mid Y,X_O,\theta^{(t)}]^TY}{2\sigma^2},
\quad
r_j=\sum_i\mathbb E[X_{ij}\mid\cdots],
\quad q_j=n-r_j.
$$
Completing the square gives
$$
Q=-(\beta-d)^TA(\beta-d)
+\sum_j\{r_j\log\pi_j+q_j\log(1-\pi_j)\}
+\text{constant}.
$$
For $v\ne0$,
$$
v^TAv=\frac1{2\sigma^2}\mathbb E\|Xv\|^2
+\frac1{2\sigma_\beta^2}\|v\|^2>0,
$$
so $A$ is positive definite. Exact rowwise expectations require summing over $2^{|U_i|}$ states. Thus the cost is exponential in the largest number of missing covariates in one row, more precisely $\sum_i2^{|U_i|}$ times a polynomial factor for accumulating first and second moments.


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 216 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-216/4/c/solution