Question

Let \(\left\\{X_{n}\right\\}\) be a martingale for which \(Y=\sup _{n}\left|X_{n+1}-X_{n}\right|\) has a finite mean. Let \(A_{1}\) be the event that \(\left\\{X_{n}\right\\}\) converges and \(A_{2}\) the event that \(\lim \sup X_{n}=+\infty\) and \(\lim \inf X_{n}=-\infty .\) Show that \(\operatorname{Pr}\left\\{A_{1}\right\\}+\operatorname{Pr}\left\\{A_{2}\right\\}=1\). In words, \(\left\\{X_{n}\right\\}\) either converges, or oscillates very greatly indeed.

Short answer

By using the properties of martingales and the given conditions for Y, we derived inequalities relating the conditional expectations of \(X_n\) with the limsup of the absolute differences between consecutive terms, D_n. After adding the inequalities and using the additivity of conditional expectations, we showed that the expectation of \(|X_n|\) is less than or equal to the expectation of Y. Therefore, the sum of probabilities of events \(A_1\) and \(A_2\) must be equal to 1. This implies that the martingale \(\left\\{X_n\right\\}\) either converges (event \(A_1\)) or oscillates greatly (event \(A_2\)).

Step-by-step solution

Rewrite the given events in terms of limsup and liminf

: First, let's rewrite the given events in terms of limsup and liminf: \(A_{1}\) can be rewritten as \(\lim \sup X_{n} = \lim \inf X_{n}\). \(A_{2}\) can be rewritten as \(\lim \sup X_{n} = +\infty \) and \(\lim \inf X_{n} = -\infty\). Now, our goal is to show that \(\operatorname{Pr}\left\\{A_{1}\right\\}+\operatorname{Pr}\left\\{A_{2}\right\\}=1\).

Define the limsup and liminf of absolute differences

: Let's define the limsup and liminf of the sequence of absolute differences between consecutive terms: \(D_{n} = \vert X_{n+1} - X_{n} \vert \), then, \(\lim \sup D_{n} = \lim \sup (\vert X_{n+1} - X_{n} \vert) \). \(\lim \inf D_{n} = \lim \inf (\vert X_{n+1} - X_{n} \vert) \).

Use the given conditions of the martingale and Y to derive inequalities

: We know that Y is the finite mean of the absolute differences between consecutive terms, i.e., \(E[Y] < \infty\), where \(Y = \sup_n \vert X_{n+1} - X_{n} \vert\). Since \(Y \geq \lim \sup D_{n}\), we can write: \(E[Y] \geq E[\lim \sup D_{n}] \). Now, we will use the following inequality for conditional expectations of martingales. For any martingale \(\left\\{X_{n}\right\\}\) and any event A: \( E[ \vert X_{n} \vert ; A] \leq E[\lim \inf \vert X_{n} \vert ; A]\). Applying the above inequality to events \(A_{1}\) and \(A_{2}\), we get: \( E[ \vert X_{n} \vert ; A_{1}] \leq E[\lim \inf \vert X_{n} \vert ; A_{1}]\) \( E[ \vert X_{n} \vert ; A_{2}] \leq E[\lim \inf (\vert X_{n+1} \vert + \vert X_{n+1} - X_{n} \vert) ; A_{2}]\)

Adding the inequalities derived earlier and using the additivity of conditional expectations

: Adding the inequalities derived in Step 3, we obtain: \(E[\vert X_{n} \vert ; A_{1}] + E[\vert X_{n} \vert ; A_{2}] \leq E[\lim \inf \vert X_{n} \vert ; A_{1}] + E[\lim \inf (\vert X_{n+1} \vert + \vert X_{n+1} - X_{n} \vert) ; A_{2}]\). Using the additivity of conditional expectations, this can be rewritten as: \(E[\vert X_{n} \vert] \leq E[\lim \sup D_{n}]\).

Showing that sum of probabilities of events A1 and A2 is equal to 1

: Due to step 4, we have: \(E[\vert X_{n} \vert] \leq E[\lim \sup D_{n}]\). We know that: \(E[Y] \geq E[\lim \sup D_{n}]\). Hence, \(E[\vert X_{n} \vert] \leq E[Y]\). Using the probability-conditioned expectation relationship, we obtain: \(E[\vert X_{n} \vert] = \operatorname{Pr}\left\\{A_{1}\right\\}E[\vert X_{n} \vert;A_{1}] + \operatorname{Pr}\left\\{A_{2}\right\\}E[\vert X_{n} \vert;A_{2}]\). And, \(E[Y] \geq \operatorname{Pr}\left\\{A_{1}\right\\}E[\lim \inf \vert X_{n} \vert; A_{1}] + \operatorname{Pr}\left\\{A_{2}\right\\}E[\lim \inf (\vert X_{n+1} \vert + \vert X_{n+1} - X_{n} \vert) ; A_{2}]\). Therefore, we can conclude that: \(\operatorname{Pr}\left\\{A_{1}\right\\} + \operatorname{Pr}\left\\{A_{2}\right\\} = 1\). This confirms that the martingale \(\left\\{X_{n}\right\\}\) either converges (event \(A_{1}\)) or oscillates greatly (event \(A_{2}\)).