By the definition of <exponential tilting>,
$$
\mathbb P(A_n)
=\mathbb E_\lambda\!\left[
e^{-\lambda S_n+n\psi(\lambda)}\mathbf1_{A_n}
\right].
$$
On $A_n$ one has $S_n\leq cn$, so for $\lambda\geq0$,
$$
e^{-\lambda S_n+n\psi(\lambda)}
\geq e^{-\lambda cn+n\psi(\lambda)}.
$$
Therefore
$$
\boxed{\mathbb P(A_n)
\geq e^{-\lambda cn+n\psi(\lambda)}
\mathbb P_\lambda(A_n).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-201/3/b/solution