Prove that $f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, -1) < B_{\tau_{n-1 }} < f_n(B_{\tau_1 } , \dots, B_{\tau_{n-1 }}, 1)$

Combining the martingale property with the inequality $f_n(x_1, \dots, x_{n-1 } , -1 ) < f_n(x_1, \dots, x_{n-1 } , 1 )$ and assuming that $\mathrm P(D_n = 1\mid \mathcal F_{n-1})\notin \{0,1\}$ a.s., we have $$ X_{n-1} = \mathrm E[X_n \mid \mathcal F_{n-1}] = \mathrm E[f_n(X_1,\dots,X_{n-1},D_n) \mid \mathcal F_{n-1}] \\ = f_n(X_1,\dots,X_{n-1},1)\cdot \mathrm P(D_n = 1\mid \mathcal F_{n-1}) \\+ f_n(X_1,\dots,X_{n-1},-1)\cdot \mathrm P(D_n = -1\mid \mathcal F_{n-1})\\ < f_n(X_1,\dots,X_{n-1},1)\cdot \mathrm P(D_n = 1\mid \mathcal F_{n-1})\\ + f_n(X_1,\dots,X_{n-1},{\color{red}1})\cdot \mathrm P(D_n = -1\mid \mathcal F_{n-1})\\ = f_n(X_1,\dots,X_{n-1},1). $$ Similarly, $$ X_{n-1} > f_n(X_1,\dots,X_{n-1},-1)\cdot \mathrm P(D_n = 1\mid \mathcal F_{n-1})\\ + f_n(X_1,\dots,X_{n-1},-1)\cdot \mathrm P(D_n = -1\mid \mathcal F_{n-1}) = f_n(X_1,\dots,X_{n-1},-1). $$

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Probability Theory

Stochastic Processes

Brownian Motion

Stopping Times

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