The <Harris-FKG inequality> gives
$$
\mathbb P(X\in A\cap B)\geq
\mathbb P(X\in A)\mathbb P(X\in B).
$$
Here is an induction proof of the product-measure version, also called <Harris' inequality>. The result is immediate for one coordinate. For $n$ coordinates, define
$$
a_j=\mathbb P(A\mid X_n=j),
\qquad
b_j=\mathbb P(B\mid X_n=j),
\qquad j\in\{0,1\}.
$$
The sections of an <increasing event> are increasing events in the first $n-1$ coordinates, so the induction hypothesis gives
$$
\mathbb P(A\cap B)
\geq(1-p_n)a_0b_0+p_na_1b_1.
$$
Monotonicity gives $a_1\geq a_0$ and $b_1\geq b_0$. The difference between the right side and
$$
\mathbb P(A)\mathbb P(B)
=\bigl((1-p_n)a_0+p_na_1\bigr)
\bigl((1-p_n)b_0+p_nb_1\bigr)
$$
is
$$
p_n(1-p_n)(a_1-a_0)(b_1-b_0)\geq0.
$$
This closes the induction and proves positive association.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 209 / 1 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-209/1/solution