Proving an inequality involving conditional probability

Question

Let $(X_t)_{t\ge0}$ be a stochastic process on a probability space $(\Omega,\mathcal F, \mathbb P)$ and let $\mathcal F_t=\sigma(X_s:0\le s\le t)$. Let $\Lambda\in \mathcal F_t$ with $\Lambda\subset \{|X_t|<a\}$ and $\mathbb P(\Lambda)>0$. Define $$ Y = \frac{1}{2 \epsilon} \int_t^{t+2\epsilon} 1_{\{x:|x|<2a\}} (X_s) \ ds. $$ How can I prove the following inequality : $$ 2\mathbb E[Y\mid\Lambda] \le 2 \mathbb P(Y>1/2\mid\Lambda) + \mathbb P(Y\le1/2\mid\Lambda) $$

Answer 1

Obviously, we have

$$0 \leq Y \leq \frac{1}{2\epsilon} \int_{t}^{t+2\epsilon} 1 \, ds = 1.$$

Hence,

$$Y = \underbrace{Y}_{\leq 1} \cdot 1_{\{Y>1/2\}} + \underbrace{Y}_{\leq 1/2} \cdot 1_{\{Y \leq 1/2\}} \leq 1_{\{Y>1/2\}} + \frac{1}{2} 1_{\{Y \geq 1/2\}}.$$

Now the claim follows by taking the conditional expectation $\mathbb{E}( \cdot \mid \Lambda)$ on both sides. (Recall that the conditional expectation is monotonic, i.e. $X \leq Z \implies \mathbb{E}(X \mid \Lambda) \leq \mathbb{E}(Z \mid \Lambda)$ for any two integrable random variables $X$, $Z$.)