Martingale Convergence¶

The Convergence Question¶

One of the most striking features of martingale theory is that martingales "want to converge." Unlike general sequences of random variables, martingales possess an intrinsic structure—the conditional expectation property—that forces limiting behavior under mild conditions.

The fundamental question: given a martingale $\{M_t\}$ or $\{M_n\}$, does $\lim M_t$ (or $\lim M_n$) exist, and in what sense?

This section develops the convergence theory, progressing from almost sure convergence to $L^p$ convergence, with the crucial role of uniform integrability as the bridge between these modes.

Almost Sure Convergence: The Upcrossing Lemma¶

The key to martingale convergence is Doob's upcrossing inequality, which bounds how often a martingale can oscillate between two levels.

Upcrossings Defined¶

For a process $\{X_n\}$ and real numbers $a < b$, an upcrossing of the interval $[a, b]$ is a pair of times $s < t$ such that $X_s \le a$ and $X_t \ge b$ (the process crosses from below $a$ to above $b$).

Let $U_N^{[a,b]}$ denote the number of upcrossings of $[a, b]$ by $(X_0, X_1, \ldots, X_N)$.

Doob's Upcrossing Inequality¶

Theorem: If $\{X_n\}_{n=0}^N$ is a submartingale, then:

\[ \boxed{(b - a) \mathbb{E}[U_N^{[a,b]}] \le \mathbb{E}[(X_N - a)^+]} \]

where $(x)^+ = \max(x, 0)$.

Proof: Define the betting strategy process $H_n$ as follows: $H_n = 1$ if we are currently "in an upcrossing" (i.e., below $a$ and waiting to cross above $b$), and $H_n = 0$ otherwise. Formally:

Let $\sigma_1 = \inf\{n \geq 0 : X_n \leq a\}$ (first time at or below $a$)
Let $\tau_1 = \inf\{n > \sigma_1 : X_n \geq b\}$ (first time at or above $b$ after $\sigma_1$)
Continue inductively: $\sigma_{k+1} = \inf\{n > \tau_k : X_n \leq a\}$, $\tau_{k+1} = \inf\{n > \sigma_{k+1} : X_n \geq b\}$

Set $H_n = 1$ if $\sigma_k \leq n - 1 < \tau_k$ for some $k$, and $H_n = 0$ otherwise. This is a predictable process (it depends only on $X_0, \ldots, X_{n-1}$).

The total gain from this strategy is:

\[ G_N = \sum_{n=1}^N H_n (X_n - X_{n-1}) \]

Each completed upcrossing contributes at least $b - a$ to the gain, so:

\[ G_N \geq (b - a) U_N^{[a,b]} - (X_N - a)^- \]

For submartingales, predictable strategies with non-negative bounded coefficients preserve the submartingale property, so $\mathbb{E}[G_N] \geq 0$. From $G_N \geq (b-a)U_N^{[a,b]} - (X_N - a)^-$ and $\mathbb{E}[G_N] \ge 0$:

\[ (b-a)\mathbb{E}[U_N^{[a,b]}] \le \mathbb{E}[(X_N - a)^-] + \mathbb{E}[G_N] \]

Now use the identity $(x-a)^+ - (x-a)^- = x - a$, so $(x-a)^- = (x-a)^+ - (x-a) \le (x-a)^+$. Combined with $\mathbb{E}[G_N] \ge 0$:

\[ (b-a)\mathbb{E}[U_N^{[a,b]}] \le \mathbb{E}[(X_N - a)^-] \le \mathbb{E}[(X_N - a)^+] \]

$\square$

Interpretation: If the expected final value is bounded, the process cannot oscillate too much. Each upcrossing "costs" at least $b - a$ on average.

The Martingale Convergence Theorem¶

Theorem (Doob's Martingale Convergence Theorem): Let $\{M_n\}_{n \ge 0}$ be a martingale (or submartingale) satisfying:

\[ \sup_n \mathbb{E}[|M_n|] < \infty \quad \text{($L^1$-bounded)} \]

Then there exists a random variable $M_\infty$ with $\mathbb{E}|M_\infty| < \infty$ such that:

\[ \boxed{M_n \to M_\infty \quad \text{almost surely as } n \to \infty} \]

Proof:

Step 1: For any rational $a < b$, the upcrossing inequality gives:

\[ \mathbb{E}[U_\infty^{[a,b]}] = \lim_N \mathbb{E}[U_N^{[a,b]}] \le \frac{\sup_n \mathbb{E}[(M_n - a)^+]}{b - a} < \infty \]

Therefore $U_\infty^{[a,b]} < \infty$ almost surely.

Step 2: The event "$(M_n)$ oscillates infinitely often" equals:

\[ \bigcup_{a < b, \, a,b \in \mathbb{Q}} \{U_\infty^{[a,b]} = \infty\} \]

This is a countable union of null sets, hence null.

Step 3: Therefore, $\liminf_n M_n = \limsup_n M_n$ almost surely, so the limit exists (possibly $\pm \infty$).

Step 4: To show $M_\infty$ is finite a.s., note that $(|M_n|)$ is a submartingale and:

\[ \mathbb{P}(|M_\infty| = \infty) \le \lim_K \mathbb{P}\left(\sup_n |M_n| \ge K\right) \le \lim_K \frac{\sup_n \mathbb{E}|M_n|}{K} = 0 \quad \square \]

L¹ Convergence and Uniform Integrability¶

Almost sure convergence does not imply $L^1$ convergence. The canonical counterexample:

Example: Let $\xi_1, \xi_2, \ldots$ be i.i.d. with $\mathbb{P}(\xi_i = 2) = \mathbb{P}(\xi_i = 0) = 1/2$. Define:

\[ M_n = \prod_{i=1}^n \xi_i \]

Then $M_n$ is a martingale with $\mathbb{E}[M_n] = 1$ for all $n$. However, $M_n \to 0$ almost surely (since $\mathbb{P}(\text{at least one } \xi_i = 0) = 1$), but $\mathbb{E}[M_n] = 1 \not\to 0 = \mathbb{E}[M_\infty]$.

Why does this happen? The expected value is preserved by increasingly rare but large realizations. With probability $(1/2)^n$, all $\xi_i = 2$, giving $M_n = 2^n$. This contributes $(1/2)^n \cdot 2^n = 1$ to the expectation—the entire expected value comes from an exponentially unlikely event.

The gap is filled by uniform integrability — the condition that rules out the probability mass escaping to infinity that we saw above. The complete theory, including three equivalent characterizations and all proofs, is developed in the companion document Uniform Integrability. We recall just the definition and key examples here.

Uniform Integrability¶

A family of random variables $\{X_\alpha\}_{\alpha \in A}$ is uniformly integrable (UI) if:

\[ \lim_{K \to \infty} \sup_{\alpha \in A} \mathbb{E}[|X_\alpha| \mathbf{1}_{\{|X_\alpha| > K\}}] = 0 \]

Interpretation: The tails of all $X_\alpha$ are controlled by the same truncation level $K$, uniformly in $\alpha$.

Key examples of UI families (see Uniform Integrability for proofs):

Any $L^p$-bounded family with $p > 1$.
The family $\{\mathbb{E}[Y \mid \mathcal{G}_\alpha]\}$ for any $Y \in L^1$ and any sub-$\sigma$-algebras $\mathcal{G}_\alpha$.
Any family dominated by an integrable random variable.

The UI Convergence Theorem¶

Theorem: Let $\{M_n\}$ be a martingale. The following are equivalent:

$(M_n)$ is uniformly integrable.
$M_n \to M_\infty$ in $L^1$ for some $M_\infty \in L^1$.
$M_n \to M_\infty$ a.s. and in $L^1$ for some $M_\infty \in L^1$.
There exists $M_\infty \in L^1$ such that $M_n = \mathbb{E}[M_\infty \mid \mathcal{F}_n]$ for all $n$.
$(M_n)$ is closed: there exists $X \in L^1$ such that $M_n = \mathbb{E}[X \mid \mathcal{F}_n]$.

Proof sketch (main implications):

$(4) \Rightarrow (1)$: Conditional expectations of an integrable random variable form a UI family.

$(1) \Rightarrow (3)$: By the basic convergence theorem, $M_n \to M_\infty$ a.s. UI plus a.s. convergence implies $L^1$ convergence (Vitali).

$(3) \Rightarrow (4)$: For any $A \in \mathcal{F}_n$, by dominated convergence:

\[ \mathbb{E}[M_\infty \mathbf{1}_A] = \lim_m \mathbb{E}[M_m \mathbf{1}_A] = \mathbb{E}[M_n \mathbf{1}_A] \]

showing $M_n = \mathbb{E}[M_\infty \mid \mathcal{F}_n]$. $\square$

L^p Convergence¶

Theorem: Let $p > 1$ and let $(M_n)$ be a martingale with $\sup_n \mathbb{E}|M_n|^p < \infty$. Then:

\[ \boxed{M_n \to M_\infty \quad \text{in } L^p \text{ and a.s.}} \]

Proof: $L^p$-boundedness with $p > 1$ implies UI, so $L^1$ (hence a.s.) convergence follows. For $L^p$ convergence, Doob's maximal inequality gives:

\[ \mathbb{E}\left[\sup_n |M_n|^p\right] \le \left(\frac{p}{p-1}\right)^p \sup_n \mathbb{E}|M_n|^p < \infty \]

Since $|M_n - M_\infty|^p \le 2^p \sup_n |M_n|^p$ which is integrable, dominated convergence yields $L^p$ convergence. $\square$

Remark: The case $p = 1$ fails: $L^1$-boundedness gives a.s. convergence but not $L^1$ convergence without UI.

Backward Martingales¶

A backward martingale (or reverse martingale) is a sequence $(M_n, \mathcal{F}_n)_{n \geq 1}$ with $\mathcal{F}_n \supseteq \mathcal{F}_{n+1}$ (decreasing filtration) and:

\[ \mathbb{E}[M_n \mid \mathcal{F}_{n+1}] = M_{n+1} \quad \text{for all } n \geq 1 \]

Equivalently, as we move forward in the index $n$, we have less information (the filtration shrinks), and conditioning on this smaller $\sigma$-algebra still gives the martingale relation.

Theorem (Backward Martingale Convergence): If $(M_n, \mathcal{F}_n)_{n \geq 1}$ is a backward martingale with $\sup_n \mathbb{E}|M_n| < \infty$, then:

\[ M_n \to M_\infty \quad \text{a.s. and in } L^1 \]

where $M_\infty = \mathbb{E}[M_1 \mid \mathcal{F}_\infty]$ and $\mathcal{F}_\infty = \bigcap_n \mathcal{F}_n$ is the tail $\sigma$-algebra.

Key difference: Backward martingales always converge in $L^1$—uniform integrability is automatic since $(M_n)$ is a sequence of conditional expectations of $M_1$.

Application (Strong Law of Large Numbers): Let $X_1, X_2, \ldots$ be i.i.d. with $\mathbb{E}|X_1| < \infty$ and mean $\mu$. Define:

$S_n = X_1 + \cdots + X_n$
$\mathcal{F}_n = \sigma(S_n, S_{n+1}, S_{n+2}, \ldots)$ — the $\sigma$-algebra generated by all partial sums from $S_n$ onward

The key observation is that $\mathcal{F}_n \supseteq \mathcal{F}_{n+1}$ (knowing $(S_n, S_{n+1}, \ldots)$ gives more information than knowing only $(S_{n+1}, S_{n+2}, \ldots)$).

Let $\bar{X}_n = S_n/n$. Then $(\bar{X}_n, \mathcal{F}_n)$ is a backward martingale. By the backward convergence theorem:

\[ \frac{S_n}{n} = \bar{X}_n \to \mathbb{E}[X_1 \mid \mathcal{F}_\infty] \quad \text{a.s. and in } L^1 \]

By Kolmogorov's 0-1 law, $\mathcal{F}_\infty$ is trivial (all events have probability 0 or 1), so:

\[ \frac{S_n}{n} \to \mathbb{E}[X_1] = \mu \quad \text{a.s. and in } L^1 \]

This elegant proof of the SLLN avoids truncation arguments.

Continuous-Time Convergence¶

The convergence theorems extend to continuous time with appropriate modifications.

Theorem: Let $(M_t)_{t \ge 0}$ be a right-continuous martingale with:

\[ \sup_t \mathbb{E}|M_t| < \infty \]

Then $M_\infty := \lim_{t \to \infty} M_t$ exists almost surely and is finite a.s.

If additionally $(M_t)$ is UI, then $M_t \to M_\infty$ in $L^1$ and $M_t = \mathbb{E}[M_\infty \mid \mathcal{F}_t]$.

Regularity: In continuous time, one typically assumes càdlàg paths (right-continuous with left limits). The Doob regularization theorem guarantees that any $L^1$-bounded martingale has a càdlàg modification.

Convergence of Specific Martingales¶

Example 1: Doob Martingale¶

If $M_t = \mathbb{E}[X \mid \mathcal{F}_t]$ for some $X \in L^1$, then $M_t \to X$ a.s. and in $L^1$ as $t \to \infty$ (assuming $\mathcal{F}_t \uparrow \mathcal{F}$).

This is the Lévy martingale convergence theorem: as information increases, our best estimate of $X$ converges to $X$ itself.

Example 2: Exponential Martingale¶

Consider $Z_t = \exp(\theta W_t - \frac{\theta^2 t}{2})$. For $\theta \neq 0$:

$\mathbb{E}[Z_t] = 1$ for all $t$.
$Z_t \to 0$ almost surely as $t \to \infty$. To see this, take logarithms: $\log Z_t = \theta W_t - \frac{\theta^2}{2} t$. By the law of the iterated logarithm, $W_t = O(\sqrt{t \log \log t})$ a.s., so $\theta W_t = o(t)$ a.s. Therefore $\log Z_t \sim -\frac{\theta^2}{2}t \to -\infty$ a.s., giving $Z_t \to 0$ a.s.

This is not UI: $\mathbb{E}[Z_t] = 1 \not\to 0 = \mathbb{E}[Z_\infty]$.

Example 3: Random Walk Martingale¶

The simple random walk $S_n = \sum_{i=1}^n \xi_i$ (fair coin) satisfies $S_n \to ?$. Since $\mathbb{E}[S_n^2] = n \to \infty$, the walk is not $L^2$-bounded, and indeed $S_n$ does not converge.

However, $S_n/\sqrt{n} \Rightarrow N(0,1)$ in distribution (CLT), though this is not a.s. convergence.

The Optional Stopping Connection¶

Martingale convergence and optional sampling are deeply connected:

Theorem: Let $(M_n)$ be a UI martingale with limit $M_\infty$. For any stopping time $\tau$ (possibly infinite):

\[ \mathbb{E}[M_\tau] = \mathbb{E}[M_0] \quad \text{and} \quad M_\tau = \mathbb{E}[M_\infty \mid \mathcal{F}_\tau] \]

This extends the optional sampling theorem to unbounded stopping times, using convergence to handle the limit.

Summary: The Convergence Hierarchy¶

Condition	Almost Sure	$L^1$	$L^p$ ($p > 1$)
$L^1$-bounded	✓	✗ (need UI)	—
Uniformly integrable	✓	✓	—
$L^p$-bounded ($p > 1$)	✓	✓	✓
Closed (= $\mathbb{E}[X \mid \mathcal{F}_n]$)	✓	✓	Depends on $X$

Key takeaways:

$L^1$-boundedness alone gives a.s. convergence via the upcrossing lemma.
Uniform integrability bridges a.s. and $L^1$ convergence and is equivalent to closure.
$L^p$-boundedness with $p > 1$ is stronger than UI and gives $L^p$ convergence.
Backward martingales always converge in $L^1$—no additional conditions needed.

The convergence theory completes the structural picture of martingales: they are processes that, under boundedness conditions, settle down to limiting values, preserving the martingale property even at "time infinity."

Exercises¶

Exercise 1: Upcrossings¶

Let $M_n$ be a submartingale and let $U_N^{[a,b]}$ denote the number of upcrossings of $[a,b]$ by $M_0, \ldots, M_N$.

(a) State Doob's upcrossing inequality.

(b) Use the upcrossing inequality to prove: if $\sup_n \mathbb{E}[M_n^+] < \infty$, then $M_n$ converges a.s.

(c) Explain why the condition involves $M_n^+$ rather than $|M_n|$.

Solution to Exercise 1

(a) Doob's upcrossing inequality: If $\{X_n\}_{n=0}^N$ is a submartingale and $a < b$, then:

\[ (b - a)\,\mathbb{E}[U_N^{[a,b]}] \le \mathbb{E}[(X_N - a)^+] \]

where $U_N^{[a,b]}$ is the number of upcrossings of $[a, b]$ by $X_0, \ldots, X_N$.

(b) For any rational $a < b$, the upcrossing inequality gives:

\[ \mathbb{E}[U_\infty^{[a,b]}] = \lim_{N \to \infty} \mathbb{E}[U_N^{[a,b]}] \le \frac{\sup_n \mathbb{E}[(M_n - a)^+]}{b - a} \]

Now $(M_n - a)^+ \le |M_n| + |a|$, so $\mathbb{E}[(M_n - a)^+] \le \mathbb{E}[|M_n|] + |a|$. For a submartingale, $\mathbb{E}[M_n^+] \le \mathbb{E}[|M_n|]$ and $\mathbb{E}[|M_n|] = 2\mathbb{E}[M_n^+] - \mathbb{E}[M_n] \le 2\mathbb{E}[M_n^+]$ (using $|x| = 2x^+ - x$). If $\sup_n \mathbb{E}[M_n^+] < \infty$, then $\sup_n \mathbb{E}[|M_n|] < \infty$.

Therefore $U_\infty^{[a,b]} < \infty$ a.s. for each rational pair $a < b$. The event "$(M_n)$ oscillates infinitely" is $\bigcup_{a < b, \, a,b \in \mathbb{Q}} \{U_\infty^{[a,b]} = \infty\}$, a countable union of null sets, hence null. So $M_n$ converges a.s. $\square$

(c) The condition involves $M_n^+$ rather than $|M_n|$ because for a submartingale, the positive part controls the negative part. Specifically, $\mathbb{E}[M_n^-] = \mathbb{E}[M_n^+] - \mathbb{E}[M_n] \le \mathbb{E}[M_n^+] - \mathbb{E}[M_0]$ (since $\mathbb{E}[M_n]$ is non-decreasing for submartingales). So bounding $\mathbb{E}[M_n^+]$ automatically bounds $\mathbb{E}[M_n^-]$ and hence $\mathbb{E}[|M_n|] = \mathbb{E}[M_n^+] + \mathbb{E}[M_n^-]$.

For a general martingale, $\mathbb{E}[M_n^+] = \mathbb{E}[M_n^-] + \mathbb{E}[M_n]$, and if $\mathbb{E}[M_n] = \mathbb{E}[M_0]$ is constant, then $\sup_n \mathbb{E}[M_n^+] < \infty$ is equivalent to $\sup_n \mathbb{E}[|M_n|] < \infty$.

Exercise 2: Uniform Integrability¶

(a) Prove that a family $\{X_\alpha\}$ is uniformly integrable if and only if it is $L^1$-bounded and satisfies: for all $\epsilon > 0$, there exists $\delta > 0$ such that $\mathbb{P}(A) < \delta$ implies $\sup_\alpha \mathbb{E}[|X_\alpha| \mathbf{1}_A] < \epsilon$.

(b) Show that $\{W_t : t \le T\}$ is uniformly integrable for any $T < \infty$.

(c) Show that $\{e^{W_t - t/2} : t \ge 0\}$ is not uniformly integrable.

Solution to Exercise 2

(a) ($\Rightarrow$) Assume $\{X_\alpha\}$ is UI. Then $\sup_\alpha \mathbb{E}[|X_\alpha|] < \infty$ (taking $K = 0$ in the definition).

For equi-absolute continuity: given $\varepsilon > 0$, choose $K$ such that $\sup_\alpha \mathbb{E}[|X_\alpha| \mathbf{1}_{\{|X_\alpha| > K\}}] < \varepsilon/2$. Then for any event $A$:

\[ \mathbb{E}[|X_\alpha| \mathbf{1}_A] = \mathbb{E}[|X_\alpha| \mathbf{1}_A \mathbf{1}_{\{|X_\alpha| \le K\}}] + \mathbb{E}[|X_\alpha| \mathbf{1}_A \mathbf{1}_{\{|X_\alpha| > K\}}] \le K\,\mathbb{P}(A) + \varepsilon/2 \]

Taking $\delta = \varepsilon/(2K)$: if $\mathbb{P}(A) < \delta$, then $\sup_\alpha \mathbb{E}[|X_\alpha| \mathbf{1}_A] < \varepsilon$.

($\Leftarrow$) Assume $L^1$-bounded and equi-absolutely continuous. Note:

\[ \mathbb{E}[|X_\alpha| \mathbf{1}_{\{|X_\alpha| > K\}}] \le \sup_\alpha \mathbb{E}[|X_\alpha| \mathbf{1}_{A_\alpha}] \]

where $A_\alpha = \{|X_\alpha| > K\}$. By Markov: $\mathbb{P}(A_\alpha) \le \mathbb{E}[|X_\alpha|]/K \le C/K$. Given $\varepsilon > 0$, choose $\delta$ from equi-absolute continuity, then $K = C/\delta$. For this $K$, $\mathbb{P}(A_\alpha) < \delta$ for all $\alpha$, so $\sup_\alpha \mathbb{E}[|X_\alpha| \mathbf{1}_{\{|X_\alpha| > K\}}] < \varepsilon$. $\square$

(b) $\sup_{t \le T} \mathbb{E}[W_t^2] = T < \infty$. Since $p = 2 > 1$ and the family is $L^2$-bounded, it is UI. $\square$

(c) As shown in Exercise 1(b): $Z_t = e^{W_t - t/2} \to 0$ a.s. but $\mathbb{E}[Z_t] = 1$, so $\mathbb{E}[Z_t] \not\to \mathbb{E}[\lim Z_t] = 0$. By Vitali's theorem, this failure of $L^1$ convergence despite a.s. convergence implies the family is not UI. $\square$

Exercise 3: Backward Martingales¶

Let $X_1, X_2, \ldots$ be i.i.d. with $\mathbb{E}[X_1] = \mu$ and $\mathbb{E}|X_1| < \infty$. Let $\bar{X}_n = \frac{1}{n}\sum_{k=1}^n X_k$.

(a) Define $\mathcal{F}_n = \sigma(S_n, S_{n+1}, \ldots)$ where $S_k = X_1 + \cdots + X_k$. Show this is a decreasing sequence of $\sigma$-algebras.

(b) Prove that $(\bar{X}_n, \mathcal{F}_n)_{n \geq 1}$ is a backward martingale.

(c) Use backward martingale convergence to prove the strong law of large numbers: $\bar{X}_n \to \mu$ a.s.

Solution to Exercise 3

(a) Define $\mathcal{F}_n = \sigma(S_n, S_{n+1}, \ldots)$ where $S_k = X_1 + \cdots + X_k$.

If we know $(S_n, S_{n+1}, S_{n+2}, \ldots)$, we can recover $(S_{n+1}, S_{n+2}, \ldots)$ (by just ignoring $S_n$). But knowing $(S_n, S_{n+1}, \ldots)$ gives strictly more information than knowing $(S_{n+1}, S_{n+2}, \ldots)$ because we also know $S_n$ and hence $X_n = S_n - S_{n-1}$ if we know $S_{n-1}$... Actually, more precisely:

$\mathcal{F}_{n+1} = \sigma(S_{n+1}, S_{n+2}, \ldots) \subseteq \sigma(S_n, S_{n+1}, S_{n+2}, \ldots) = \mathcal{F}_n$

since every function of $(S_{n+1}, S_{n+2}, \ldots)$ is also a function of $(S_n, S_{n+1}, S_{n+2}, \ldots)$. So $\mathcal{F}_n \supseteq \mathcal{F}_{n+1}$: the sequence is decreasing. $\square$

(b) We need to show $\mathbb{E}[\bar{X}_n \mid \mathcal{F}_{n+1}] = \bar{X}_{n+1}$ for all $n \ge 1$.

Since $\bar{X}_n = S_n/n$ and $S_n = S_{n+1} - X_{n+1}$: $\bar{X}_n = (S_{n+1} - X_{n+1})/n$.

Given $\mathcal{F}_{n+1} = \sigma(S_{n+1}, S_{n+2}, \ldots)$, $S_{n+1}$ is $\mathcal{F}_{n+1}$-measurable. We need $\mathbb{E}[X_{n+1} \mid \mathcal{F}_{n+1}]$. By the exchangeability argument: given $(S_{n+1}, S_{n+2}, \ldots)$, the individual summands $X_1, \ldots, X_{n+1}$ are exchangeable (by the i.i.d. assumption). Therefore:

\[ \mathbb{E}[X_{n+1} \mid \mathcal{F}_{n+1}] = \mathbb{E}[X_k \mid \mathcal{F}_{n+1}] \quad \text{for all } k = 1, \ldots, n+1 \]

Summing over $k$: $(n+1)\mathbb{E}[X_{n+1} \mid \mathcal{F}_{n+1}] = \mathbb{E}[S_{n+1} \mid \mathcal{F}_{n+1}] = S_{n+1}$.

So $\mathbb{E}[X_{n+1} \mid \mathcal{F}_{n+1}] = S_{n+1}/(n+1) = \bar{X}_{n+1}$.

Therefore:

\[ \mathbb{E}[\bar{X}_n \mid \mathcal{F}_{n+1}] = \frac{1}{n}\mathbb{E}[S_{n+1} - X_{n+1} \mid \mathcal{F}_{n+1}] = \frac{S_{n+1} - S_{n+1}/(n+1)}{n} = \frac{S_{n+1} \cdot n/(n+1)}{n} = \frac{S_{n+1}}{n+1} = \bar{X}_{n+1} \]

So $(\bar{X}_n, \mathcal{F}_n)$ is a backward martingale. $\square$

(c) By the backward martingale convergence theorem, $\bar{X}_n \to M_\infty$ a.s. and in $L^1$, where $M_\infty = \mathbb{E}[X_1 \mid \mathcal{F}_\infty]$ and $\mathcal{F}_\infty = \bigcap_{n \ge 1} \mathcal{F}_n$ is the tail $\sigma$-algebra.

By Kolmogorov's zero-one law, $\mathcal{F}_\infty$ is trivial (all events have probability 0 or 1). This is because $\mathcal{F}_\infty \subseteq \sigma(X_{n+1}, X_{n+2}, \ldots)$ for all $n$ (the tail $\sigma$-algebra of the partial sums is contained in the tail $\sigma$-algebra of the i.i.d. sequence), and the latter is trivial by Kolmogorov's 0-1 law.

Since $\mathcal{F}_\infty$ is trivial, $\mathbb{E}[X_1 \mid \mathcal{F}_\infty] = \mathbb{E}[X_1] = \mu$ a.s. Therefore:

\[ \bar{X}_n = \frac{S_n}{n} \to \mu \quad \text{a.s.} \quad \square \]

Condition	Almost Sure	\(L^1\)	\(L^p\) (\(p > 1\))
\(L^1\)-bounded	✓	✗ (need UI)	—
Uniformly integrable	✓	✓	—
\(L^p\)-bounded (\(p > 1\))	✓	✓	✓
Closed (= \(\mathbb{E}[X \mid \mathcal{F}_n]\))	✓	✓	Depends on \(X\)