How to get $\mathbb E[a^{\tau_1} \phi(X_{\tau_1}) | X_0 =x] = \mathbb E[a^{\tau_2} \phi(X_{\tau_2}) | X_0 =x]$ from Strong Markov property?

Question

Consider a continuous-time Markov chain $(X_t)_{t \ge 0}$ with respect to a completed right-continuous filtration $(\mathcal G_t)_{t \ge 0}$. Suppose that

• The state space $V$ is finite and endowed with discrete topology.

• $a \in (0,1)$ and $\phi$ is a measurable function from $V$ to $\mathbb R_+$.

• $\tau_1 \le \tau_2$ are stopping times.

Then my professor said that by Strong Markov property, we have $$\mathbb E[a^{\tau_1} \phi(X_{\tau_1}) | X_0 =x] = \mathbb E[a^{\tau_2} \phi(X_{\tau_2}) | X_0 =x]$$

---

Could you please elaborate on how to get the above equality from this version of Strong Markov property?

Answer 1

Thank you so much for @Saad's comment. I post it here to close this question.

---

This assertion seems incorrect. E.g. if $τ_1=0$ and $τ_2=1$, it asserts that$$φ(x)=E_x(a^{τ_1}φ(X_{τ_1}))=E_x(a^{τ_2}φ(X_{τ_2}))=aE_x(φ(X_1)),\quad\forall a\in(0,1),\ x\in V$$ which implies that $φ=0$.