Question

Let \(X, X_{1}, X_{2}, \ldots\) be independent identically distributed random varialhi'm having negative mean \(\mu\) and finite variance \(\sigma^{2}\). With \(S_{0}=0\) and \(S_{n}=X_{1}\) ? \(\cdots+X_{n}\), set \(M=\max _{n \geq 0} S_{n} .\) In view of \(\mu<0\), we know that \(M<\infty .\) Assumi \(E[M]<\infty\). (In fact, it can be shown that this is a consequence of \(\sigma^{2}<\infty\).) Define \(r(x)=x^{+}=\max \\{x, 0\\}\) and \(f(x)=E\left[(x+M-E[M])^{+}\right]\) (a) Show \(f(x) \geq r(x)\) for all \(x\). (b) Show \(f(x) \geq E[f(x+X)]\) for all \(x\), so that \(\left\\{f\left(x+S_{n}\right)\right\\}\) is a nonnegative supermartingale [Hint: Verify and use the fact that \(M\) and \((X+M)^{+}\)have the same distribution.] (c) Use (a) and (b) to show \(f(x) \geq E\left[\left(x+S_{T}\right)^{+}\right]\)for all Markov times \(T .\left[\left(x+S_{\infty}\right)^{+}=\lim _{n \rightarrow \infty}\left(X+S_{n}\right)^{+}=0 .\right]\)

Short answer

For this question, we first derive some facts about the maximum sum M of the random variables, and then we proceed to answer each part: (a) We show that if x ≥ 0, the inequality is trivial, and for x < 0, we have f(x) ≥ r(x). (b) Using the same distribution among M', M, and X, we prove f(x) ≥ E[f(x + X)] for all x, resulting in a non-negative supermartingale. (c) Applying the Optional Stopping Theorem and considering any Markov stopping time T, we prove the inequality f(x) ≥ E[(x + Sₜ)⁺] for all Markov times T.

Step-by-step solution

Part (a): Proving f(x) ≥ r(x) for all x.

We first notice that if x ≥ 0, the inequality is trivial because we have f(x) = E[(x + M - E[M])⁺] ≥ E[x] = x, so our main concern lies with x < 0. For x < 0, we have: f(x) = E[(x + M - E[M])⁺] ≥ E[(M - E[M])⁺], since adding any negative value to E[(M - E[M])⁺] will result in a larger or equal right-hand side. Now, consider the random variable M'. We define M' = M - E[M] + x. Moreover, x + M - E[M] = x + (M - E[M]), which tells us that the random variable (x + M - E[M])⁺ follows the same distribution as (M')⁺. Knowing that, we have: f(x) = E[(x + M - E[M])⁺] = E[(M')⁺] ≥ E[(M - E[M])⁺] = r(x), as needed.

Part (b): Proving f(x) ≥ E[f(x + X)] for all x.

We start by noticing that M itself follows the same distribution as (X + M)⁺, as hinted. Therefore, M and (X + M)⁺ have the same mean and variance. To prove the inequality f(x) ≥ E[f(x + X)], let us compute E[f(x + X)]: E[f(x + X)] = E[E[(x + X + M - E[M])⁺]] = E[E[(X + (x + M - E[M]))⁺]]. Now, let's apply the conditional expectation to compute the inner expectation: E[(X + (x + M - E[M]))⁺ | M] = E[(X + M')⁺ | M], since M' = x + M - E[M] as we defined previously. We know that X + M follows the same distribution as M by the hint, so X + M' follows the same distribution as M': E[(X + M')⁺ | M] = E[(M')⁺] = f(x). Using the tower property of conditional expectation, we have: E[f(x + X)] = E[E[f(x + X) | M]] = E[f(x)], which implies: f(x) ≥ E[f(x + X)]. Moreover, from part (a), we know that f(x) is non-negative. Hence, we can conclude that {f(x + Sₙ)} is a non-negative supermartingale.

Part (c): Proving f(x) ≥ E[(x + Sₜ)⁺] for all Markov times T.

Using the Optional Stopping Theorem for supermartingales, if τ is a stopping time such that E[τ] is finite, we have: E[f(x + Sₜ)] ≤ f(x). Now, let T be any Markov stopping time. Since f(x) is non-negative and E[f(x + Sₜ)] ≤ f(x), we have that E[f(x + Sₜ) - (x + Sₜ)⁺] ≤ 0, which implies: f(x) ≥ E[f(x + Sₜ)] ≥ E[(x + Sₜ)⁺]. The inequality holds for all Markov times T, as required.

Key concepts

Supermartingale

In the realm of stochastic processes, a supermartingale is a pivotal concept that bears an innate relation to martingales, but with a slight twist. Imagine a gambler whose fortune over time forms a sequence, with each step depending on some randomness. A martingale would then represent a fair game for this gambler, where his expected future fortune, given all past fortunes, remains the same. However, for a supermartingale, this expected future fortune is not more than the current fortune, hinting at a game that is somehow 'unfair' or disadvantageous over time.

A supermartingale can be formally defined as a sequence of random variables \( \{X_n\} \), such that for every \( n \), the conditional expectation \( E[X_{n+1} | \mathcal{F}_n] \leq X_n \), where \( \mathcal{F}_n \) represents the filtration up to time \( n \). The step by step exercise walks through proving that a particular sequence forms a non-negative supermartingale, illuminating this concept with a real-world application. The process is supermartingale if it shows a tendency to decrease on average but allows for random fluctuations at each step, capturing the essence of risk in stochastic processes.

Conditional Expectation

Conditional expectation lies at the heart of understanding stochastic processes. It's akin to having a crystal ball that provides a glimpse into the future, but only with the knowledge of the present and past. Mathematically, it is defined as the expected value of a random variable given some condition, often related to a certain amount of information available at the time.

More formally, the conditional expectation \( E[X | \mathcal{F}] \) is the expectation of the random variable \( X \) given the sigma-algebra \( \mathcal{F} \), which encapsulates the information up to that point. In the exercise provided, conditional expectations play a central role in demonstrating properties of the function \( f(x) \). To understand the behavior of \( f(x) \), it is necessary to compute expectations based on the information at hand. The concept serves as a bridge between what is known (the conditioning information) and what is sought (the expectation of some random outcome), reflecting the impact of accumulating information on the prediction of future events.

Markov Times

The concept of Markov times, also known as stopping times, is intriguing as it introduces the element of decision-making into the world of stochastic processes. Suppose you're watching a sports match, and you decide to leave when the first goal is scored. That moment when you leave can be considered a Markov time because it's a point in the future that you can't predict with certainty, but when it happens, you know it immediately.

More technically, a Markov time with respect to a filtration \( \mathcal{F}_n \) is a random variable \( T \) that takes non-negative integer values and satisfies the condition that for every \( n \), the event \( \{ T \leq n \} \) is in \( \mathcal{F}_n \). This means you can determine whether the stopping time has occurred or not using only the information available up to time \( n \). In the step by step solution, the concept of Markov times is essential in part (c), as it applies the Optional Stopping Theorem to show that the inequality holds for all such stopping times, thus underscoring the integral role of timing in stochastic analysis.

Optional Stopping Theorem

The Optional Stopping Theorem is a striking result in the study of stochastic processes and it provides significant insight into the behavior of martingales and supermartingales at random times. Imagine being in a game wherein you have the option to stop playing at any moment. This theorem essentially tells you what to expect if you stop your game based on some strategy or rule that doesn't look into the future, that is, a Markov time.

The Optional Stopping Theorem states that, given certain conditions, the expected value of a supermartingale at a stopping time is less than or equal to its expected value at the start. Formally, if \( X_n \) is a supermartingale and \( T \) is a Markov time such that certain conditions are met (like \( T \) having a finite expectation), then \( E[X_T] \leq E[X_0] \). This theorem is artfully utilized in part (c) of the solution to deduce the expected value of the function \( f(x) \) at the stopping time, cementing the Optional Stopping Theorem's role as a crucial tool for analyzing the stopping behavior of stochastic sequences.