 Multiplicative upcrossing supermartingale

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.