Let
$$
S_n=\sum_{i=1}^nX_i^{(n)}.
$$
Here $S_n$ has the <binomial distribution> with parameters $(n,\lambda/n)$. Part a, now with independent coordinates, gives
$$
D_e(P_{S_n}\Vert\operatorname{Poisson}(\lambda))
\leq n(\lambda/n)^2=\frac{\lambda^2}{n}.
$$
By <Pinsker's inequality>, the probability mass functions therefore converge in total variation, and in particular $S_n$ <convergence in distribution>[converges in distribution] to $Z\sim\operatorname{Poisson}(\lambda)$.

The joint probability of the observed row depends only on $S_n$:
$$
P_n(X_1^{(n)},\ldots,X_n^{(n)})
=\left(\frac\lambda n\right)^{S_n}
\left(1-\frac\lambda n\right)^{n-S_n}.
$$
Taking logarithms in any fixed base and choosing $c_n=\log n$ gives
$$
-\frac1{c_n}\log P_n(X_1^{(n)},\ldots,X_n^{(n)})
=a_nS_n+b_n,
$$
where
$$
a_n=\frac{\log(n/\lambda)+\log(1-\lambda/n)}{\log n}\longrightarrow1,
\qquad
b_n=-\frac{n\log(1-\lambda/n)}{\log n}\longrightarrow0.
$$
The convergence lemma supplied in the question now yields
$$
-\frac1{\log n}\log P_n(X_1^{(n)},\ldots,X_n^{(n)})
\xrightarrow{d}Z,
\qquad Z\sim\operatorname{Poisson}(\lambda).
$$
This sparse triangular array therefore has a random limiting normalized self-information rather than the constant limit in the usual <asymptotic equipartition property>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 224 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-224/4/b/solution