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[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{equation} \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} \end{equation} Let $Y_n=1$ for $0\leq n<N_1$ and for $j\geq 1$ \begin{equation} Y_n=\begin{cases}(b/a)^{j-1}(X_n/a)&\text{for}N_{2j-1}\leq n<N_{2j},\\(b/a)^j&\text{for}N_{2j}\leq n<N_{2j+1}\end{cases} \end{equation} 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{equation} \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<n)}\text{is a supermartingale.}\end{aligned} \end{equation}

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?

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1 Answer 1

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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.

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