• $(E_i,\mathcal E_i)$ be a measurable space;
  • $f:E_1\times E_2\to E_1\to E_1$ be $(\mathcal E_1\otimes\mathcal E_2,\mathcal E_1)$-measurable;
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
  • $X_0$ be an $E_1$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$;
  • $(Z_n)_{n\in\mathbb N}$ be an $E_2$-valued i.i.d. process on $(\Omega,\mathcal A,\operatorname P)$ independent of $X_0$ and $\mu$ denote the distribution of $Z_1$.

How do we show that $$X_n:=f(X_{n-1},Z_n)\;\;\;\text{for }n\in\mathbb N$$ is a time-homogeneous Markov chain with transition kernel $$(\kappa g)(x):=\operatorname E\left[(g\circ f)(x,Z_1)\right]=\int(g\circ f)(x,z)\:\mu({\rm d}z)\;\;\;\text{for }x\in E_1$$ for bounded $\mathcal E_1$-measurable $g:E_1\to\mathbb R$?

Remember the following result:

Lemma: Let $Y_i$ be an $(E_i,\mathcal E_i)$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$, $h:E_1\times E_2\to\mathbb R$ is bounded and $(\mathcal E_1\otimes\mathcal E_2)$-measurable and $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra. If $\mathcal F$ is independent of $Y_2$ and $Y_1$ is $\mathcal F$-measurable, then $$\operatorname E\left[h(Y_1,Y_2)\mid\mathcal F\right]=\left.\operatorname E\left[h(y_1,Y_2)\right]\right|_{y_1\:=\:Y_1}\tag1.$$

Using this lemma, it is easy to see that $$\operatorname E\left[g(X_{n+1})\mid\mathcal F^X_n\right]=(\kappa g)(X_n)\tag2$$ holds true for $n=0$.

But how do we need to proceed for $n\ge1$? Are we somehow able to show that $\mathcal F^X_n$ is independent of $Z$?

  • 1
    $\begingroup$ Notice that $\sigma(X_0, \dots, X_{n-1}) \subseteq \sigma(X_0, Z_1, \dots, Z_{n-1})$ which by assumption is independent of $\sigma(Z_n, Z_{n+1}, \dots)$, and in particular is independent of $Z_n$. $\endgroup$ Apr 10, 2022 at 18:07
  • $\begingroup$ @NateEldredge Thank you. It's trivial. That was the easy argument I was looking for. $\endgroup$
    – 0xbadf00d
    Apr 10, 2022 at 18:31

1 Answer 1


For convenicen Let $Z'_n=Z_{n-1}$ for $n\geq1$ so that $X_n(\omega)=f(X_{n-1}(\omega),Z'_{n-1}(\omega))$

It suffices to show that for any measurable $A\subset E_1$, $$E[\mathbb{1}_A(X_{n+1})|\sigma(X_0,\ldots,X_n)=G_n(X_n)$$ for each $n\geq1$.

It is easy to see that there are measurable functions $F_n:E_1\times E^n_2\rightarrow E_1$ such that $$X_n=F_n(X_0,Z'_0,\ldots,Z'_{n-1})$$ Fix $n\in\mathbb{N}$. Any bounded measurable function $H(\omega):=h(X_0(\omega),\ldots,X_n(\omega))$ depends on $X_0,\ldots,X_n$ only through $(X_0,Z'_1,\ldots,Z'_{n-1})$, that is $H=\Phi(X_0,Z'_0,\ldots,Z'_{n-1})$. As $X_0$ and $Z'$ are independent

$$\begin{align} E[\mathbb{1}_A((X_{n+1})H]&=E[\mathbb{1}_A(F_n(X_0,Z'_0,\ldots,Z'_n))\Phi(X_0,Z'_0,\ldots,Z'_{n-1})]\\ &=\int_{E_1\times E^{n+1}_2 }\mathbb{1}_A(F_n(x_0,z_0,\ldots,z_{n-1},z_n))\Phi(x_0,z_0,\ldots,z_{n-1})\mu_{X_0}(dx_0)\mu^{\otimes n+1}_{Z'_0}(dz_0,\ldots,dz_n)\\ &=\int_{E_0\times E^n_2}\Phi(x_0,z_0,\ldots,z_{n-1})\cdot\\ &\qquad\Big(\int_{E_2}\mathbb{1}_A(F_n(x_0,z_0,\ldots,z_{n-1},z_n))\,\mu_{Z_0}(dz_n)\Big)\,\mu_{X_0}(dx_0)\times\mu^{\otimes n}_{Z_0}(dz_0,\ldots,dz_n)\\ &=\int\Phi(x_0,z_0,\ldots,z_{n-1})\Big(\int\mathbb{1}_A(f(x_n,z_n)\,\mu_{Z_0}(dz_n)\Big)\mu_{X_0}(dx_0)\times\mu^{\otimes n}_{Z_0}(dz_0,\ldots,dz_n)\\ &=E[HG_n(X_n)] \end{align} $$ where $G(x_n)=\int\mathbb{1}_A(f(x_n,z)\mu_{Z_0}(dz)$. That is, $$E[\mathbb{1}_A(X_{n+1})|\sigma(X_0,X_1,\ldots, X_n)]=G(X_n)$$


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