Martingale Stopped Values: Uniform Integrability Proof

Hey guys! Today, we're diving deep into the fascinating world of martingales, specifically focusing on proving that the stopped values of a martingale are uniformly integrable. This is a crucial concept in stochastic processes, especially when dealing with convergence theorems and applications in areas like mathematical finance. So, buckle up, and let's get started!

Understanding Martingales and Uniform Integrability

Before we jump into the proof, let's make sure we're all on the same page regarding the key concepts. A martingale is, simply put, a stochastic process that, on average, stays the same over time. More formally, a stochastic process $(M_t)_{t \geq 0}$ is a martingale with respect to a filtration $(\mathcal{F}_t)_{t \geq 0}$ if:

1. $M_t$ is $\mathcal{F}_t$-measurable for all $t$.
2. $E[|M_t|] < \infty$ for all $t$.
3. $E[M_{t+s} | \mathcal{F}_t] = M_t$ for all $t, s \geq 0$.

In simpler terms, the best prediction for the future value of the martingale, given all the information we have up to time $t$, is just its current value at time $t$.

Now, what about uniform integrability? A family of random variables $(X_i)_{i \in I}$ is said to be uniformly integrable if for every $\epsilon > 0$, there exists a $K > 0$ such that:

$\sup_{i \in I} E[|X_i| \cdot \mathbb{1}_{|X_i| > K}] < \epsilon$

Intuitively, uniform integrability means that the tails of the distributions of the random variables in the family are uniformly small. This is a stronger condition than just having the random variables be bounded in $L^1$.

Why is uniform integrability important? Well, it plays a crucial role in ensuring that convergence in $L^1$ implies convergence of expectations. Specifically, if $X_n \to X$ in $L^1$ and the sequence $(X_n)$ is uniformly integrable, then $E[X_n] \to E[X]$. This is essential for many applications, especially when dealing with limits of stochastic processes.

The Stopped Values of a Martingale

Now, let's introduce the concept of a stopping time. A stopping time $\tau$ with respect to a filtration $(\mathcal{F}_t)_{t \geq 0}$ is a random variable taking values in $[0, \infty]$ such that for every $t \geq 0$, the event ${\tau \leq t}$ is in $\mathcal{F}_t$. In other words, whether or not the stopping time has occurred by time $t$ is determined by the information available up to time $t$.

The stopped value of a martingale $M_t$ at a stopping time $\tau$ is simply $M_{\tau}$. More formally, we define:

$M_{\tau} = M_{\tau(\omega)}(\omega)$

where $\omega$ represents a sample path. Stopped martingales are incredibly useful for analyzing the behavior of martingales at random times. They appear frequently in option pricing theory, sequential analysis, and other areas of stochastic modeling.

Proving Uniform Integrability of Stopped Values

Okay, let's get to the heart of the matter: proving that the stopped values of a martingale are uniformly integrable. Suppose $(M_t)_{t \geq 0}$ is a martingale and $\tau$ is a stopping time. We want to show that the family of random variables $(M_{\tau} \mathbb{1}_{\tau < \infty})$ is uniformly integrable. We will assume that $M_0 = 0$ to simplify the calculations (otherwise consider $M_t - M_0$).

Here's the general idea of the proof:

1. Relate the stopped value to the maximum of the martingale: Show that $E[|M_{\tau}|] \leq 2 E[\sup_{t \geq 0} |M_t|]$.
2. Use Doob's maximal inequality: This inequality provides a bound on the expected value of the supremum of the martingale. Specifically, for any $c > 0$, $P(\sup_{t \geq 0} |M_t| > c) \leq \frac{E[|M_{\infty}|]}{c}$, where $M_{\infty}$ is the limit of the martingale (assuming it converges).
3. Apply the dominated convergence theorem: Show that as $c \to \infty$, the tails of the distribution of $|M_{\tau}|$ become uniformly small.

Now, let's dive into the details:

Let $A \in \mathcal{F}_\infty$. We have:

$E[M_{\tau} \mathbb{1}_A] = E[\sum_{t=0}^{\infty} M_t \mathbb{1}_{\tau = t} \mathbb{1}_A] = \sum_{t=0}^{\infty} E[M_t \mathbb{1}_{\tau = t} \mathbb{1}_A] = \sum_{t=0}^{\infty} E[E[M_{\infty} | \mathcal{F}_t] \mathbb{1}_{\tau = t} \mathbb{1}_A] = \sum_{t=0}^{\infty} E[M_{\infty} \mathbb{1}_{\tau = t} \mathbb{1}_A] = E[M_{\infty} \mathbb{1}_A \mathbb{1}_{\tau < \infty}]$

Therefore, $E[M_{\tau} | \mathcal{F}_\infty] = M_{\infty} \mathbb{1}_{\tau < \infty}$. This implies that $M_{\tau} \mathbb{1}_{\tau < \infty} = E[M_{\infty} | \mathcal{F}_\infty] \mathbb{1}_{\tau < \infty}$, and thus $E[|M_{\tau}|] \leq E[|M_{\infty}|]$.

Now, let $K > 0$. We have:

$E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > K}] = E[|E[M_{\infty} | \mathcal{F}_\infty]| \mathbb{1}_{|E[M_{\infty} | \mathcal{F}_\infty]| > K}] \leq E[E[|M_{\infty}| | \mathcal{F}_\infty] \mathbb{1}_{|E[M_{\infty} | \mathcal{F}_\infty]| > K}] = E[|M_{\infty}| \mathbb{1}_{|E[M_{\infty} | \mathcal{F}_\infty]| > K}]$

Since $E[|M_{\infty}|] < \infty$, we know that $M_{\infty}$ is integrable. As $K \to \infty$, $E[|M_{\infty}| \mathbb{1}_{|M_{\infty}| > K}] \to 0$. Also, since $|E[M_{\infty} | \mathcal{F}_\infty]| \leq E[|M_{\infty}| | \mathcal{F}_\infty]$, if $|E[M_{\infty} | \mathcal{F}_\infty]| > K$, then $|M_{\infty}|$ must be large enough, so the expectation goes to zero.

By the dominated convergence theorem, as $K \to \infty$, $E[|M_{\infty}| \mathbb{1}_{|E[M_{\infty} | \mathcal{F}_\infty]| > K}] \to 0$. Therefore, for any $\epsilon > 0$, there exists a $K > 0$ such that $E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > K}] < \epsilon$. This proves that the stopped values of the martingale are uniformly integrable.

An Alternative Proof

Here's another way to approach the proof, which relies on Doob's maximal inequality more explicitly.

Let $M_t^* = \sup_{0 \leq s \leq t} |M_s|$. By Doob's maximal inequality, for any $c > 0$:

$P(M_t^* > c) \leq \frac{E[|M_t|]}{c}$

Since $M_t$ is a martingale, $E[|M_t|] = E[|M_0|]$, which is constant. Let's assume $M_0 = 0$. Thus, $E[|M_t|] = 0$ and $P(M_t^* > c) = 0$ for all $c > 0$.

Now, consider $E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > c}]$. We can write this as:

$E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > c}] = E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > c, \tau \leq t}] + E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > c, \tau > t}]$

For the first term:

$E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > c, \tau \leq t}] \leq E[M_t^* \mathbb{1}_{M_t^* > c}] \to 0$ as $c \to \infty$ because $M_t^*$ is integrable due to Doob's $L^p$ maximal inequality.

For the second term, we know that $P(\tau > t) \to 0$ as $t \to \infty$. Therefore, $E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > c, \tau > t}] \leq E[|M_{\tau}| \mathbb{1}_{\tau > t}] \to 0$ as $t \to \infty$.

Combining these results, we have:

$\lim_{c \to \infty} \sup_{\tau} E[|M_{\tau}| \mathbb{1}_{|M_{\tau}| > c}] = 0$

This confirms that the stopped values of the martingale are uniformly integrable.

Why This Matters

Understanding that the stopped values of a martingale are uniformly integrable is not just an abstract theoretical result. It has significant practical implications. For example, in mathematical finance, this result is crucial for justifying the convergence of certain numerical methods used to price options. It also plays a key role in proving the convergence of stochastic algorithms used in machine learning and optimization.

Conclusion

So, there you have it! We've successfully navigated through the proof that the stopped values of a martingale are uniformly integrable. This involves understanding the definitions of martingales, stopping times, and uniform integrability, as well as leveraging powerful tools like Doob's maximal inequality and the dominated convergence theorem. Hopefully, this explanation has shed some light on this important concept and its applications. Keep exploring, and happy learning!