In Durrett's textbook, the strong Markov property is defined as follows:

For every bounded and measurable $\varphi$ and stopping time $N$: $$\mathbb{E}(\varphi\circ\theta_N|\mathcal{F}_N)=\mathbb{E}_{X_N}(\varphi) \tag{1}$$

But I think the strong markov property should be: $$\mathbb{E}(\varphi(X_N,X_{N+1},\cdots)|\mathcal{F}_N)=\mathbb{E}(\varphi(X_N,X_{N+1},\cdots)|X_N) \tag{2}$$ for every bounded and measurable $\varphi$.

I want to deduce (2) from (1).


First I suppose $\varphi=\varphi(\omega_0,\omega_1,\cdots)$ in (1)

then $$\mathbb{E}_{X_N}(\varphi) =\sum_{k=0}^\infty\mathbb{E}(\varphi(\omega_0,\omega_1,\cdots)|X_0=k)1_{\{X_N=k\}}\overset{\text{time homogeneous}}{=}\sum_{k=0}^\infty\mathbb{E}(\varphi(\omega_N,\omega_{N+1},\cdots)|X_N=k)1_{\{X_N=k\}}=\mathbb{E}(\varphi(\omega_N,\omega_{N+1},\cdots)|X_N)$$ so we have: $$\mathbb{E}(\varphi(\omega_N,\omega_{N+1},\cdots)|\mathcal{F}_N)=\mathbb{E}(\varphi(\omega_N,\omega_{N+1},\cdots)|X_N)$$

the reasoning above is right or not ?

  • $\begingroup$ Use \tag{1} for equation numbers. $\endgroup$
    – user147263
    Dec 5, 2014 at 20:40
  • $\begingroup$ Hello. How do you take the first equality in the update? Thanks $\endgroup$
    – perlman
    Nov 27, 2017 at 3:15

1 Answer 1


Hint: Take the conditional expectation (with respect to the $\sigma$-algebra generated by $X_N$) in $(1)$ and use the tower property.

  • $\begingroup$ I updated my post,I didn't take the conditional expectation,is there something wrong? $\endgroup$
    – Lookout
    Dec 6, 2014 at 1:37
  • $\begingroup$ @user166445 Looks fine... $\endgroup$
    – saz
    Dec 6, 2014 at 8:58

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