Inequality about integral of probability density function in minimax theory

Question

Recently I have been reading some material about minimax theory and there is an inequality that I don't understand how to derive it. That is \begin{align*} \int[p_0^n(x)\land p_1^n(x)]dx \geq\frac{1}{2}\left(1-\frac{1}{2}\int|p_0-p_1|\right)^{2n}, \end{align*} here $p_0$ and $p_1$ are two probability density functions. I know that since $$ p_0(x)\land p_1(x):=\min(p_0(x),p_1(x))=\frac{1}{2}(p_0(x)+p_1(x)-|p_0(x)-p_1(x)|), $$ the above inequality is equivalent to $$ \int[p_0^n(x)\land p_1^n(x)]dx \geq\frac{1}{2}\left(\int[p_0(x)\land p_1(x)]dx\right)^{2n}. $$ Then I am not sure how to get the desired result. Any advice would be appreciated!

Answer 1

I have deduced a method that may be a little complicated. First, we can prove that $$ \int p \land q \geq \frac{1}{2}e^{-KL(P,Q)}, $$ where $p$ and $q$ are p.d.f.s of the distribution $P$ and $Q$. It can be done by observing that $\int p\vee q+\int p\land q=2$ and applying Jensen's inequality. Then, with another observation that $KL(P^n,Q^n)=nKL(P,Q)$, we only need to prove that $$ e^{-nKL(P,Q)}\geq(1-\frac{1}{2}\int|p-q|)^{2n}, $$ which is equivalent to $$ e^{-KL(P,Q)} \geq (1-\frac{1}{2}\int|p-q|)^2. $$ It can be done by applying the Bretagnolle–Huber inequality with $d_{TV}(P,Q)=\frac{1}{2}\int|p-q|$.