Stopping time and its probability showing $\mathbb{P}(\exists n \in \mathbb{N}_{0} : \sum_{k=1}^{n}V_{k} > a) \leq e^{-\lambda a}$

Question

The problem is related (or based on) Wald's Equation, but I am struggling finishing the proof.

We have $(V_{k})_{k \in \mathbb{N}}$ i.i.d. random Variables on the Probability-space $(\Omega, F, \mathbb{P})$ with $\mathbb{E}[e^{\lambda V_{1}}] \leq 1$ for $\lambda > 0$ and $(F_{n})_{n \in \mathbb{N}}$ being the standard filtration.

Now there are two parts of the exercise:

(a) For $a>0$ show that $\tau := \inf\{n \in \mathbb{N} | \sum_{k=1}^{n}V_{k} > a\}$ is a stopping time

(b) Show that: $\mathbb{P}(\exists n \in \mathbb{N}_{0} : \sum_{k=1}^{n}V_{k} > a) \leq e^{-\lambda a}$

Proof

(a) I need to show that $\{\tau \leq n \} \in F_{n}$ $\forall n \in \mathbb{N} \cup \{\infty\}$ . First of all I focus on the sum and rewrite it as: $$\sum_{k=1}^{n}V_{k} > a \iff \sum_{k=1}^{\infty}V_{k}1_{\{n> k-1 \}} > a$$ since $\{n> k-1 \}$ only depends on $\{V_{1},..,V_{k-1}\}$ at most. It follows: $\{V_{1},..,V_{k-1}\} \in F_{k-1} \subset F_{n}$ . But I don't know how to formalize the conclusion and properly write it down? $$\{\tau \leq n \} = \{\inf\{n \in \mathbb{N} | \sum_{k=1}^{n}V_{k} > a\} \leq n \} = \{\inf\{n \in \mathbb{N} | \sum_{k=1}^{\infty}V_{k}1_{\{n> k-1 \}} > a\} \leq n \} \in F_{n}$$

(b) There exists $n \in \mathbb{N}_{0}$ and set $S_{n} := \sum_{k=1}^{n}V_{k}$ such that using the Markov Inequality: $$\mathbb{P}( S_{n} > a) \leq \frac{\mathbb{E}[S_{n}]}{a}$$ So now I need to show that $\mathbb{E}[S_{n}] \leq e^{\lambda}$. Analog to the proof of Walds Equation I derive: $$\mathbb{E}[S_{n}] \leq \mathbb{E}[V_{1}] \mathbb{E}[\tau]$$

This is what I came up with but at this point I am stuck. $\mathbb{E}[e^{\lambda V_{1}}] \leq 1$ looks like the moment generating function but I don't know if I can follow $\mathbb{E}[V_{1}] \leq 1$ from that since I don't have a density function given.

Thanks in advance

Answer 1

I finally found an answer to (b) in the original question. Foolishly I didn't consider all the data from the original question which I hereby add:

in a previous problem I showed $X_{n} := \prod_{k=1}^{n}e^{\lambda V_{k}}$ being a supermartingale. And it was given that $X_{0} := 1$

$$\mathbb{P}(\sum_{k=1}^{n} V_{k} > a) = \mathbb{P}(X_{\tau} > e^{\lambda a}) \leq \frac{\mathbb{E}[X_{\tau}]}{e^{\lambda a}} \leq \frac{\mathbb{E}[X_{0}]}{e^{\lambda a}} = \frac{1}{e^{\lambda a}}$$

I am using the Markov Inequality and optimal stopping theorem, which I also described previously in my comments. There have been computational mistakes in my comments which prevented me finding the solution on an otherwise correct idea.

Answer 2

I think this has nothing to do with Wald's equation, but rather with optional stopping. For $N\ge 1$, denote $\tau_N = \tau\wedge N$. Then, $$ e^{\lambda a}\cdot \mathbb{P}(\tau_N<N)\le \mathbb{E}[e^{\lambda S_{\tau_N}}] \leq 1 $$ Now let $N\to\infty$.