A question about Markov Chains and transition matrix

Question

Let $X_{0}$ a random variable that takes values in the space $E$, where $E$ is countable and let $\{Y_{n}\}_{n}$ a succession of random variables independent and identically distributed $Unif(0,1)$. Suppose that there is a function $G:W \times [0,1] \rightarrow E$ and define $X_{n+1}=G(X_{n},Y_{n+1})$, proof that $\{X_{n}\}$ it's a Markov Chain and represent the transition matrix $P$ in terms of the function $G$. I've been trying to do this exercise for a while, now I know that since $X_{n+1}=G(X_{n},Y_{n+1})$ where $Y_{n+1}$ is independent, we have:\begin{align}P(X_{n+1}|\{X_n\}_{n})&=P(G(X_n,Y_{n+1})|\{X_n\}_{n})\\&=P(G(X_n,Y_{n+1})|X_n)\\&=P(X_{n+1}|X_n)\end{align} which tell us that $\{X_{n}\}$ is a Markov Chain, but I don´t know what to do in order to have that transition matrix $P$, thanks for your help.

Answer 1

Let $i$ and $j$ be elements of $E$. Because $Y_{n+1}$ is independent of $X_n$, $$\begin{align*} P(X_{n+1} = j \, | \, X_n = i) &= P(G(X_n, Y_{n+1}) = j \, | \, X_n = i) \\ &= P(G(i,Y_{n+1}) = j \, | \, X_n = i) \\ &= P(G(i,Y_{n+1}) = j) . \end{align*}$$ All the random variables $Y_{n+1}$ have the same distribution, so this last probability does not depend on $n$. Then $P$ is the matrix whose entry $(i,j)$ is the probability above.