If we have two conditions $$M_n(\hat\theta_n)+o_p(1)\ge M_n(\theta_0)$$ and $$M_n(\theta_0)\overset{P}\to M(\theta_0).$$
Can we obtain the result below directly $$M_n(\hat\theta_n)+o_p(1)\ge M(\theta_0)?$$
2026-04-22 10:37:11.1776854231
A question about a probability inequality
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$\def\th{\theta}$ $\def\ze{\zeta}$
The first condition means that there exist random variable $\xi_n$ with $\xi_n\to0$ in probability such that $$ M_n(\hat\th_0) + \xi_n \ge M_n(\th_0). $$ Let $\ze_n=M_n(\th_0)-M(\th_0)$. Then $\ze_n\to0$ in probability and $$ M_n(\hat\th_0) + \xi_n - \ze_n \ge M(\th_0). $$ Since $\xi_n-\ze_n\to0$ in probability, this means $$ M_n(\hat\th_0) + o_p(1) \ge M(\th_0). $$