====== Upcrossing Inequality and Martingale Convergence ======

===== Doob's Upcrossing Inequality =====

Let $a<b$ and let $(X_k)_{k\ge0}$ be a real-valued process.  
We define an indicator process $(C_k)$ that tracks the phases during which the process is performing an upcrossing from $a$ to $b$.

Set
$$
C_0 := \mathbf 1_{\{X_0<a\}}.
$$
For all $k\ge1$, define recursively
$$
C_k
=
\mathbf 1_{\{C_{k-1}=1\}}\mathbf 1_{\{X_{k-1}\le b\}}
+
\mathbf 1_{\{C_{k-1}=0\}}\mathbf 1_{\{X_{k-1}<a\}}.
$$

Define also
$$
Y_k := \sum_{\ell=1}^k C_\ell (X_\ell - X_{\ell-1}).
$$

**Interpretation.**
  * $C_k=1$ indicates that the process is currently in an upcrossing phase, started after going below $a$.
  * Once the process exceeds $b$, the upcrossing phase ends and $C_k$ returns to $0$.
  * The process $Y_k$ accumulates only the increments of $X$ occurring during these upcrossing phases.

<WRAP center round tip 80%>
Let $U_{0:n}[a,b]$ denote the number of completed upcrossings from $a$ to $b$ between times $0$ and $n$.  
Then the following inequality holds:
$$
Y_n \ge (b-a)\,U_{0:n}[a,b] - (X_n-a)^-.
$$

</WRAP>

===== Proof =====

Each completed upcrossing from $a$ to $b$ contributes at least $b-a$ to the sum $Y_n$, since only the increments occurring during the corresponding phase are accumulated.

The only possible negative contribution comes from an unfinished upcrossing at time $n$, when $X_n<a$.  
This loss is exactly controlled by the term $(X_n-a)^-$.

Summing the contributions of all completed upcrossings and accounting for the final correction yields the stated inequality. ∎

===== Theorem 1 (Doob's Forward Convergence Theorem) =====

<WRAP center round todo 80%>
Let $(X_n)_{n\ge0}$ be a supermartingale such that
$$
\sup_n \mathbb E[|X_n|] < \infty.
$$
Then the limit
$$
\lim_{n\to\infty} X_n
$$
exists almost surely.

</WRAP>

===== Proof =====

By the upcrossing inequality, for any $a<b$,
$$
(b-a)\,\mathbb E[U_{0:n}[a,b]]
\le
\mathbb E[Y_n] + \mathbb E[(X_n-a)^-].
$$

Since $(X_n)$ is a supermartingale, $(Y_n)$ is also a supermartingale with non-positive expectation, hence $\mathbb E[Y_n]\le0$.  
The $L^1$ boundedness assumption implies that $\sup_n \mathbb E[(X_n-a)^-]<\infty$.

By the monotone convergence theorem, 
$$
\mathbb E[U_{0:\infty}[a,b]]<\infty,
$$
which implies that the total number of upcrossings is almost surely finite.  
Hence, the process cannot oscillate infinitely many times between $a$ and $b$, and $X_n$ converges almost surely. ∎

====== Lévy's Downward Theorem ======

**Intuition.**  
Conditioning with respect to smaller and smaller $\sigma$-fields means losing information.  
The conditional expectations stabilize and converge to the conditional expectation with respect to the remaining common information.

===== Theorem 2 (Lévy's Downward Convergence Theorem) =====

<WRAP center round todo 80%>
Let $(\mathcal F_{-n})_{n\ge1}$ be a decreasing sequence of $\sigma$-fields such that
$$
\mathcal F_{-\infty}
=
\bigcap_{k\ge1}\mathcal F_{-k}
\subseteq \cdots \subseteq \mathcal F_{-(n+1)} \subseteq \mathcal F_{-n} \subseteq \cdots \subseteq \mathcal F_{-1}.
$$

For any random variable $Z$ satisfying $\mathbb E[|Z|]<\infty$, we have
$$
\lim_{n\to\infty}\mathbb E[Z\mid\mathcal F_{-n}]
=
\mathbb E[Z\mid\mathcal F_{-\infty}]
\quad\text{almost surely}.
$$

</WRAP>

===== Proof =====

Define the process
$$
X_k := \mathbb E[Z\mid\mathcal F_k], \qquad -n\le k\le -1.
$$
This is a martingale bounded in $L^1$.

Similarly as before, we define $C_1=\mathsf{1}_{X_{-n} < a}$ . For all $-n  < k \leq -1$, we set $C_k=\mathsf{1}_{C_{k-1}=1} \mathsf{1}_{X_{k-1} \leq b}+ \mathsf{1}_{C_{k-1}=0} \mathsf{1}_{X_{k-1} <a}$. Define also $Y_k=\sum_{\ell=-n}^k C_\ell (X_\ell - X_{\ell-1})$. Then, for the same reasons, the number of upcrossing $U_{-n:-1}[a,b]$ between $-n$ and $-1$ can be bounded as follows:  
$$
Y_{-1} \geq (b-a) U_{-n:-1}[a,b] - (X_{-1}-a)^-
$$

As before, the integrability of $Z$ ensures that the expected number of upcrossings is finite, which implies that $(X_k)$ converges almost surely as $k\to-\infty$.

The limit must coincide with $\mathbb E[Z\mid\mathcal F_{-\infty}]$, which completes the proof. ∎

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