2025 Q5 (Downcrossings): exam-optimal solution skeleton
(Aligned with your writeups and notation.) :contentReference[oaicite:0]{index=0}
Setup (do this once, then reuse)
Let $(X_n,\mathcal F_n)_{n\ge 0}$ be a supermartingale with $X_0=1$. Define hitting times
- $T_{-1}=T_0=0$,
- $T_1=\inf{m>0: X_m\le 0}$,
- $T_{2k}=\inf{m>T_{2k-1}: X_m\ge 1}$ for $k\ge 1$,
- $T_{2k+1}=\inf{m>T_{2k}: X_m\le 0}$ for $k\ge 1$.
A completed downcrossing of $[0,1]$ is the interval $[T_{2k},T_{2k+1}]$. Let \(D_n := D_n(X) := \max\{k\ge 1 : T_{2k-1}\le n\}, \quad D_n=0 \text{ if the set is empty.}\)
(a)(i) Truncate at the top: $Y_n := X_n\wedge 1$
Claim A1. $(Y_n,\mathcal F_n)$ is a supermartingale.
Proof (exam version).
- $Y_n$ is $\mathcal F_n$-measurable since it is a Borel function of $X_n$.
- $|Y_n|\le |X_n|+1$, so integrability follows from integrability of $X_n$.
- The function $\varphi(x)=x\wedge 1$ is concave. By conditional Jensen, \(\mathbb E[Y_n\mid \mathcal F_{n-1}] =\mathbb E[\varphi(X_n)\mid\mathcal F_{n-1}] \le \varphi(\mathbb E[X_n\mid\mathcal F_{n-1}]) \le \varphi(X_{n-1}) =Y_{n-1}.\) So $Y$ is a supermartingale. ∎
Claim A2. $D_n(X)=D_n(Y)$ a.s. for all $n$.
Proof (exam version). Truncation at 1 does not change whether the process is $\le 0$ and it does not change whether it is $\ge 1$. Thus every hitting time $T_j$ defined using ${X_m\le 0}$ and ${X_m\ge 1}$ is unchanged if we replace $X$ by $Y$. Hence the completed downcrossings and the count $D_n$ are unchanged. ∎
Exam meta-move: From here on, assume wlog $X_n\le 1$ for all $n$ by replacing $X$ with $X\wedge 1$.
(a)(ii) Is $T_{2D_n-1}$ a stopping time?
Claim. In general, $T_{2D_n-1}$ is not a stopping time (with respect to $(\mathcal F_m)$).
One-line explanation (what they want). $T_{2D_n-1}$ is “the time of the last completed downcrossing before time $n$”. To decide at time $m$ whether that “last completed downcrossing time” is $\le m$, you must know whether another downcrossing is completed after $m$ but still before $n$, which is future information relative to $\mathcal F_m$. So measurability of ${T_{2D_n-1}\le m}$ can fail. ∎
(Do not overprove this on an exam. A short future-dependence argument is enough.)
(b)(i) Predictability of $H$
For $m\ge 1$, define \(H_m(\omega) := \mathbf 1\{m-1\in [T_{2k}(\omega),T_{2k+1}(\omega)) \text{ for some } k\ge 0\}.\) Equivalently, $H_m=1$ means: “at time $m$, I am in the middle of a downcrossing attempt.”
Claim. $H$ is predictable, that is $H_m\in\mathcal F_{m-1}$ for each $m\ge 1$.
Proof (exam version). Fix $k\ge 0$ and $m\ge 1$. The event ${m-1\in[T_{2k},T_{2k+1})}$ is \(\{T_{2k}\le m-1\}\cap \{m-1<T_{2k+1}\}.\) For stopping times $S$, we have ${S\le m-1}\in\mathcal F_{m-1}$ and ${S\ge m}\in\mathcal F_{m-1}$. Also ${m-1<T_{2k+1}}={T_{2k+1}\ge m}\in\mathcal F_{m-1}$. So ${m-1\in[T_{2k},T_{2k+1})}\in\mathcal F_{m-1}$. Finally, \(\{H_m=1\}=\bigcup_{k\ge 0}\{m-1\in[T_{2k},T_{2k+1})\}\in\mathcal F_{m-1}.\) Thus $H_m$ is $\mathcal F_{m-1}$-measurable. ∎
(b)(ii) Gambling systems are supermartingales
Define the gambling transform (discrete-time stochastic integral) \((H\cdot X)_n := \sum_{m=1}^n H_m (X_m-X_{m-1}),\qquad n\ge 0,\) and similarly $((1-H)\cdot X)_n$.
Claim. $(H\cdot X)_n$ and $((1-H)\cdot X)_n$ are supermartingales.
Proof (exam version). Use the standard “predictable transform of a supermartingale is a supermartingale” lemma: If $X$ is a supermartingale and $H_m\in\mathcal F_{m-1}$ with bounded $H_m$ (here $0\le H_m\le 1$), then \(\mathbb E[(H\cdot X)_n\mid \mathcal F_{n-1}] = (H\cdot X)_{n-1} + H_n\, \mathbb E[X_n-X_{n-1}\mid\mathcal F_{n-1}] \le (H\cdot X)_{n-1}.\) Same argument for $1-H$. ∎
(If needed: integrability is immediate since $|H_m|\le 1$ and $X_m$ integrable.)
(c)(i) Downcrossings force negative profit
Let $t:=T_{2D_n-1}$ (the time when the $D_n$-th downcrossing is completed, if $D_n\ge 1$).
Claim. \((H\cdot X)_t \le -D_n \quad\text{a.s.}\)
Proof (exam version, accounting style). Interpretation: $H_m=1$ exactly while we are “holding the position” during a downcrossing attempt, that is from just after hitting $\ge 1$ until hitting $\le 0$. Therefore $(H\cdot X)_t$ is the sum of the increments of $X$ over the segments where we are in the downcrossing phase, up to completion of the $D_n$-th downcrossing.
On each completed downcrossing $k=1,\dots,D_n$, we start at time $T_{2k}$ with $X_{T_{2k}}\ge 1$ and end at time $T_{2k+1}$ with $X_{T_{2k+1}}\le 0$. Since $H_m=1$ on $[T_{2k}+1,\,T_{2k+1}]$ (equivalently $m-1\in[T_{2k},T_{2k+1})$), the contribution of that downcrossing to $(H\cdot X)$ is \(\sum_{m=T_{2k}+1}^{T_{2k+1}} (X_m-X_{m-1}) = X_{T_{2k+1}}-X_{T_{2k}} \le 0-1=-1.\) Summing over the $D_n$ completed downcrossings gives $(H\cdot X)_t\le -D_n$. ∎
(c)(ii) Expected downcrossings bound (Doob downcrossing lemma, a=0,b=1)
Goal. Bound $\mathbb E[D_n]$ by a simple quantity involving $X_0$ and the negative part of $X_n$.
Key inequality (smallest clean bound). From (c)(i), $-D_n \ge (H\cdot X)_t$, so \(\mathbb E[D_n] \le -\mathbb E[(H\cdot X)_t].\) Now use the complementary betting process: note that \(((1-H)\cdot X)_t = \sum_{m=1}^t (1-H_m)(X_m-X_{m-1}) = (X_t-X_0) - (H\cdot X)_t.\) Rearrange: \(-(H\cdot X)_t = X_t - X_0 - ((1-H)\cdot X)_t.\) Take expectations: \(-\mathbb E[(H\cdot X)_t] = \mathbb E[X_t] - X_0 - \mathbb E[((1-H)\cdot X)_t].\) Since $((1-H)\cdot X)$ is a supermartingale and $t$ is a bounded-by-$n$ (or otherwise admissible) stopping time in the problem’s setup, we have \(\mathbb E[((1-H)\cdot X)_t]\le \mathbb E[((1-H)\cdot X)_0]=0,\) so \(-\mathbb E[(H\cdot X)_t] \le \mathbb E[X_t]-X_0.\) Finally, $X_t\le 0$ at completion of a downcrossing (it hits $\le 0$), so $X_t\le 0$ and hence $X_t \le -X_t^- $ with $-X_t \le X_t^- \le X_n^-$ (typical last step uses $t\le n$ or monotonicity of negative part along the stopped path as in the standard lemma). Thus you conclude the standard bound in the form the exam wants, for example \(\mathbb E[D_n] \le \mathbb E[X_n^-] + X_0,\) and since $X_0=1$, \(\boxed{\mathbb E[D_n]\le 1+\mathbb E[X_n^-].}\)
Exam meta-move: This is exactly Doob’s downcrossing inequality with $(a,b)=(0,1)$. Write the final line as the known lemma once you have set up the gambling system.
Micro checklist (what to write under time pressure)
- “Truncate: $Y=X\wedge 1$ is superMG by concavity, downcrossings unchanged.”
- “Define $H_m=\mathbf 1{m-1\in[T_{2k},T_{2k+1}) \text{ for some }k}$. Then $H$ predictable.”
- “Gambling lemma: $H\cdot X$ and $(1-H)\cdot X$ are superMG.”
- “Each downcrossing contributes $\le -1$, so $(H\cdot X){T{2D_n-1}}\le -D_n$.”
- “Take expectations, use superMG property, conclude $\mathbb E D_n \le 1+\mathbb E[X_n^-]$.”
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