Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(X_n)_{n\in\mathbb N}$ a martingale w.r.t. the filtration $(\mathcal F_n)$. Let $N$ a stopping time on $\mathbb N$. Prove that $(X_{N\wedge n})_n$ is a martingale w.r.t. $(\mathcal F_n)$.
Attempts
I have to prove that $$\mathbb E[X_{N\wedge (n+1)}\mid \mathcal F_n]=X_{N\wedge n}.$$
So let $F\in \mathcal F_n$. Then, $$\mathbb E[\mathbb E[X_{N\wedge (n+1)}\mid \mathcal F_n]\boldsymbol 1_F]=\mathbb E[\boldsymbol 1_F X_{N\wedge (n+1)}]=\sum_{p=0}^\infty\mathbb E[\boldsymbol 1_FX_{N\wedge (n+1)}\mid N=p]\mathbb P\{N=p\}.$$
Q1) Is it true that $$\mathbb E[\boldsymbol 1_FX_{N\wedge (n+1)}\mid N=p]=\mathbb E[\boldsymbol 1_FX_{p\wedge (n+1)}]\ \ ?$$
If yes, then $$\sum_{p=0}^\infty\mathbb E[\boldsymbol 1_FX_{N\wedge (n+1)}\mid N=p]\mathbb P\{N=p\}=\sum_{o=0}^\infty \mathbb E[\boldsymbol 1_FX_{p\nabla (n+1)}]\mathbb P\{N=p\}=\sum_{p=0}^n\mathbb E[\boldsymbol 1_FX_p]\mathbb P\{N=p\}+\sum_{k=n+1}^\infty \mathbb E[\boldsymbol 1_FX_{n+1}]\mathbb P\{N=p\}.$$
Since $(X_n)$ is a martingale, $$\mathbb E[\boldsymbol 1_FX_{n+1}]=\mathbb E[\boldsymbol 1_F\mathbb E[X_{n+1}\mid \mathcal F_n]]=\mathbb E[\boldsymbol 1_FX_n]$$
Q2) How can I continue ?
Best Answer
Note that $$ \begin{aligned} E(X_{N\wedge n+1}| \mathcal F_n) &= E(X_{N} 1_{N\leq n}+X_{n+1} 1_{N\geq n+1}| \mathcal F_n)\\ &= 1_{N\leq n} E(X_{N} | \mathcal F_n) + 1_{N\geq n+1}E(X_{n+1}| \mathcal F_n)\\ &= 1_{N\leq n} X_N + 1_{N\geq n+1}X_n\\ &= X_{N\wedge n} \end{aligned}$$