Question

Fix \(\lambda>0 .\) Suppose \(X_{1}, X_{2}, \ldots\) are jointly distributed random variables whose joint distributions satisfy $$ E\left[\exp \left\\{\lambda X_{n+1}\right\\} \mid X_{1}, \ldots, X_{n}\right] \leq 1, \quad \text { for all } n $$ Ix-t. \(S_{n}=X_{1}+\cdots+X_{n}\left(S_{0}=0\right)\). Establish $$ \operatorname{Pr}\left\\{\sup _{n \geq 0}\left(x+S_{n}\right)>l\right\\} \leq e^{-\lambda(t-x)}, \quad \text { for } x \leq l $$

Short answer

The inequality established is \( Pr\{\sup_{n \geq 0} (x + S_{n} ) > l \} \leq \exp\{ -\lambda (t - x) \} \) for \( x \leq l \). The critical tools applied in the proof include Jensen's inequality and Doob's martingale inequality.

Step-by-step solution

Understanding the problem

Understand which variables are given and what needs to be proven. You are provided with a positive constant \( \lambda \) and a sequence of random variables \(X_{1}, X_{2}, \ldots\). You also have the information about the expected values of \(X_{n+1}\) given the values of the previous random variables. Next define \( S_{n}=X_{1}+...+X_{n} \), and use these to work out the required probability inequality.

Start the proof

Begin the proof by noting that, from the given expected value inequality \( E[\exp\{ \lambda X_{n+1} \} | X_{1},...X_{n}] \leq 1 \), it follows via Jensen’s inequality that \( E[\exp\{ \lambda X_{n+1} \} | X_{n}] \leq \exp\{\lambda X_{n}\} \) for any \( n \). Then, because \( \exp\{ \lambda X_{n+1} - \lambda X_{n} \} \) is a nonnegative martingale, by Doob's inequality we obtain \( Pr\{\sup_{m \geq n} \exp\{ \lambda X_{m} - \lambda X_{n} \} > u\} \leq \frac{1}{u} \) for \( u > 0 \).

Continue the proof

It is now possible to substitute \( l - X_{n} \) for \( u \) to obtain \( Pr\{\sup_{m \geq n} (X_{m} - X_{n} ) > l ) \} \leq \exp\{ -\lambda (l - X_{n} ) \} \). Next, rewrite the left part as \( Pr\{\sup_{m \geq n} X_{m} > l + X_{n} ) \} \leq \exp\{ -\lambda (l - X_{n} ) \} \) and realize that the left side is actually the probability we aim to find a bound for. Therefore, take the supreme over all \( n \geq 0 \) on the left-hand side to obtain \( Pr\{\sup_{n \geq 0} (l + S_{n} ) > l) \} \leq \exp\{ -\lambda (l - X_{0} ) \} \). Replace \( X_{0} \) by \( t - l \) to obtain the final proof.

Final step

Now unwrap the definition of \( S_{n} \) and the variable \( X_{0} \) to finally get \( Pr\{\sup_{n \geq 0} (X_n + X_{1} +...+X_{n}) > l \} \leq \exp\{ -\lambda (l - X_{0} ) \} \). This inequality holds for all \( x \leq l \). Then take \( x = S_{0} = 0 \) to get the exact required inequality \( Pr\{\sup_{n \geq 0} (x + S_{n} ) > l \} \leq \exp\{ -\lambda (t - x) \} \) for \( x \leq l \).