= Multiplicative upcrossing supermartingale
{title2=$q=b/a$}
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>.
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