# Durrett Exercise 4.2.10 (i)

[Durrett Exercise 4.2.10 (i)] Let $$(X_n,\mathcal{F}_n)$$ be a supermartingale. Let $$N_0=-1$$ and for $$j\geq 1$$, let \begin{aligned}N_{2j-1}&=\inf\{m>N_{2j-2}:X_m\leq a\},\\N_{2j}&=\inf\{m>N_{2j-1}:X_m\geq b\}.\end{aligned} Let $$Y_n=1$$ for $$0\leq n and for $$j\geq 1$$ $$$$Y_n=\begin{cases}(b/a)^{j-1}(X_n/a)&\text{for}N_{2j-1}\leq n Use the switch principle and induction to show $$Z_n^j=Y_{n\wedge N_j}$$ is a supermartingale.

[Switching Principle] If $$X_n,Y_n$$ are supermartingales, and $$N$$ a stopping time with $$X^1_N\geq X^2_N$$, then \begin{aligned}&X_n^11_{(N>n)}+X_n^21_{(N\leq n)}\text{is a supermartingale,}\\&X_n^11_{(N\geq n)}+X_n^21_{(N

This post How to prove Dubin's inequality? gave a proof of this exercise. Following its route,

\begin{aligned} Z^1_n&=1_{(N_1>n)}+\left(\frac{X_{N_1}}{a}\right)1_{(N_1\leq n)}\\ Z^2_n&=Z^1_n1_{(N_1>n)}+\left(\frac{X_n}{a}\right)1_{(N_1\geq n)}1_{(N_2>n)}+\left(\frac{b}{a}\right)^21_{(N_2\leq n)}\\ Z^3_n&=Z^2_n1_{(N_3>n)}+\left(\frac{b}{a}\frac{X_{N_3}}{a}\right)1_{(N_3\leq n)}\\ \dots \end{aligned}

If I understand correctly, there's no reason for $$\left(\frac{X_{N_{2j-1}}}{a}\right)$$ to be supermartingales and hence to use the switching principle, since Is $X_N$ a martingale if $N$ is a stopping time? is false in general. Is there any other ways to solve this exercise?

Note $$X_N 1_{(N \leq n)} = X_{N\wedge n} 1_{(N \leq n)}$$ and if $$X_n$$ is a supermartingale and $$N$$ is a stopping time, then $$X_{N\wedge n}$$ is a supermartingale.