Let
- $(E,\mathcal E)$ be a measurable space
- $\mu$ be a probability measure on $(E,\mathcal E)$
- $X_n$ be an $(E,\mathcal E)$-valued random variable for $n\in\mathbb N$
- $\kappa$ be a Markov kernel on $(E,\mathcal E)$
- $p\in[0,1]$
Assume we construct a sequence $(Y_n)_{n\in\mathbb N}$ of $(E,\mathcal E)$-valued random variables in the following way: With probability $p$ we draw $Y_n$ from $\mu$ for all $n\in\mathbb N$ and with probability $1-p$ we draw $Y_n$ from $\kappa(X_n,\;\cdot\;)$ for all $n\in\mathbb N$.
I'm confused whether $(Y_n)_{n\in\mathbb N}$ given $(X_n)_{n\in\mathbb N}$ is distributed according to $$E^{\mathbb N}\times\mathcal E^{\otimes\mathbb N}\ni(x,B)\mapsto\bigotimes_{n\in\mathbb N}\left(p\mu+(1-p)\kappa(x_n,\;\cdot\;)\right)\tag1$$ or $$E^{\mathbb N}\times\mathcal E^{\otimes\mathbb N}\ni(x,B)\mapsto p\mu^{\otimes\mathbb N}+(1-p)\bigotimes_{n\in\mathbb N}\kappa(x_n,\;\cdot\;).\tag2$$
Intuitively, it should be $(2)$, but formally it actually seems to be $(1)$. What is correct?
Moreover, I'd be interested in the following: If $\mu$ and $\kappa$ admit densities $f$ and $g$ with respect to a common reference measure $\lambda$, do $(1)$ and $(2)$ admit a density with respect to $\lambda^{\otimes\mathbb N}$? The infinite product seems to be problematic. But is there a density at least when we replace the infinite product with a finite one?
You take the product of measures when you have independence, so $(1)$ would be the conditioned law of $(Z_n)_{n\in \mathbb{N}}$ where you draw independtly for each $Z_i$ whether or not they follow the law $\mu$ with probability $p$. The conditioned distribution of your sequence $(Y_n)_{n\in \mathbb{N}}$ would be $(2)$ assuming the $(X_n)_{n\in \mathbb{N}}$ are independent.