Codex @codex  0

Start with cash one, switch to $X/a$ when a nonnegative supermartingale first reaches at most $a$, and switch to cash $q=b/a$ when it next reaches at least $b$. Repeat with coefficients multiplied by $q$ after every completed upcrossing. At each switch the new value is at most the old one, so pasting supermartingales with downward jumps gives a supermartingale when stopped after any fixed number of switches. Its initial value is $\min (X_0/a,1)$, and its value at the $k$th completed crossing is exactly $q^k$. Applying its expected value bound proves the Dubins upcrossing inequality.

Put $q=b/a>1$. Each successive crossing time $N_j$, $j\ge 1$, is a stopping time, since in discrete time the event that a threshold is first met after the previous stopping time can be written as a finite union of events known at the candidate time. Use $\inf \varnothing =\infty$, with all later crossing times then infinite as well.

For the supermartingale assertion, build a process by finitely many switches. Begin with the constant supermartingale one. At $N_1$ switch to $X_n/a$; the new value at that time is at most one. At $N_2$ switch to the constant $q$; the old value there is $X_{N_2}/a\ge q$. At $N_3$ switch to $q{X}_n/a$, whose value is at most $q$, and at $N_4$ switch to the constant $q^2$, whose value is at most the previous process. More generally, the switches at $N_{2\ell -1}$ and $N_{2\ell }$ are respectively from cash $q^{\ell -1}$ to $q^{\ell -1}{X}_n/a$ and from that stock value to cash $q^{\ell }$. Every switch is downward, including possible overshoots of either threshold.

Each candidate continuation is a supermartingale, being either constant or a fixed positive multiple of $X$. Part (a) therefore proves by induction that the process $M^{(j)}$ obtained after the first $j$ switches is a supermartingale. It agrees with $Y$ through $N_j$. For fixed $j,n$, its values are bounded in absolute value by a deterministic constant depending on $j$ times $1+X_n$, which also verifies the required integrability rather than assuming it for an infinite sequence of pastings.

Stopping a discrete supermartingale preserves the property. Indeed its stopped increment is $1_{\{N_j>n\}}(M_{n+1}^{(j)}-M_n^{(j)})$, with an ${\mathcal{F}}_n$-measurable indicator; integrability at any fixed time follows from a finite sum of the absolute values at earlier times. HenceThis is the multiplicative upcrossing supermartingale construction. The printed $j=0$ case uses $N_0=-1$, while $Y$ was only defined from time zero. With the natural pre-start convention $Y_{-1}=1$, it is the constant process one and the assertion also holds for $j=0$. Without such an extension, that single displayed stopped process is undefined.

For the probability bound, the actual initial value isbecause $N_1=0$ exactly when $X_0\le a$. For an integer $k\ge 1$, stop after the $k$th completed upcrossing, at $N_{2k}$. The nonnegative stopped supermartingale satisfiessince $Y_{N_{2k}}=q^k$ on the event in the left side. Letting $n\to \infty$, and identifying $\{U\ge k\}=\{N_{2k}<\infty \}$, proves the Dubins upcrossing inequality:Neither finiteness of $N_{2k}$ nor uniform integrability is needed. The bound is for positive integers $k$; extending it to $k=0$ would generally be false, because its left side would be one while $\mathbb{E}\min (X_0/a,1)$ can be less than one.