Show that : $ \mathbb{P}[\max_{i=1}^{n}{X_i}\ge \epsilon]\leq\frac {1}{\epsilon}\int_{[\max_{i=1}^{n}{X_i}\ge \epsilon]} X_n d\mathbb{P} $

Question

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\left(X_n\right)_{n\ge 1} $ is a sequence of real random variables in $L^{1}(\Omega,\mathcal{F},\mathbb{P})$ and $\left(\mathcal{F}_n\right)_{n\ge 1} $ is an increasing sequence of sub- $\sigma$ -algebra of $\mathcal{F}$ . such as :

$(a)$. $\forall n\ge 1$, $X_n $ is $\mathcal{F}_n$-measurable function.

$(b)$. $\forall n\ge 1$, $\mathbb{E}[X_{n+1}\mid \mathcal{F}_n]\ge X_n ~~a.s.$

Show that, $\forall \epsilon > 0$ and $\forall n\ge 1$ :

$$ \mathbb{P}[\max_{i=1}^{n}{X_i}\ge \epsilon]\leq\frac {1}{\epsilon}\int_{[\max_{i=1}^{n}{X_i}\ge \epsilon]} X_n d\mathbb{P} $$

Answer 1

Using induction, for $n=1$ is trivial for markov inequality. We note that $ X_{n} 1_{\{ \underset{i=1,...,n}\max X_i \geq \epsilon \}} = X_n 1_{\{ \underset{i=1,...,n - 1}\max X_i \geq \epsilon \}}1_{\{ X_n < \epsilon \}} + X_n 1_{\{ X_n \geq \epsilon \}}$ and $\mathbb{E}(X_{n} 1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}} )\geq\mathbb{E}(X_{n-1}1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}})\geq \epsilon\mathbb{E}(1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}})$, for condition b) and the induction hypothesis. Note that $X_{n} 1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}} = X_n 1_{\{ \underset{i=1,...,n - 1}\max X_i \geq \epsilon \}}1_{\{ X_n < \epsilon \}} + X_n 1_{\{ \underset{i=1,...,n - 1}\max X_i \geq \epsilon \}}1_{\{ X_n \geq \epsilon \}}\Rightarrow \mathbb{E}(X_n 1_{\{ \underset{i=1,...,n - 1}\max X_i \geq \epsilon \}}1_{\{ X_n < \epsilon \}})\geq \epsilon\mathbb{E}(1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}}) - \mathbb{E}(X_n 1_{\{ \underset{i=1,...,n - 1}\max X_i \geq \epsilon \}}1_{\{ X_n \geq \epsilon \}})$, finally \begin{eqnarray} \mathbb{E}(X_{n} 1_{\{ \underset{i=1,...,n}\max X_i \geq \epsilon \}}) &\geq& \mathbb{E}(X_n 1_{\{ X_n \geq \epsilon \}}) + \epsilon\mathbb{E}(1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}}) - \mathbb{E}(X_n 1_{\{ \underset{i=1,...,n - 1}\max X_i \geq \epsilon \}}1_{\{ X_n \geq \epsilon \}})\\ & = & \epsilon\mathbb{E}(1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}}) + \mathbb{E}(X_n 1_{\{ \underset{i=1,...,n - 1}\max X_i < \epsilon \}}1_{\{ X_n \geq \epsilon \}})\geq \epsilon\mathbb{E}(1_{\{ \underset{i=1,...,n-1}\max X_i \geq \epsilon \}}) \end{eqnarray} and we conclude