Let $\sigma>0$, $\varphi(x):=\frac1{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$ for $x\in\mathbb R$ and $\psi(x):=\sum_{k\in\mathbb Z}\varphi(k+x)$ for $x\in(-1,1)$, $p\in[0,1]$, $m(x):=p+(1-p)\psi(x)$ for $x\in(-1,1)$ and $q(x):=\prod_{n\in\mathbb N}m(x_n)$ for $x\in(-1,1)^{\mathbb N}$.
I'm interested on the wrapped normal distribution with variance $\sigma^2$ on $[0,1)$. The corresponding Markov kernel can be defined as follows: Let $$\iota:\mathbb R\to[0,1)\;,\;\;\;x\mapsto x-\lfloor x\rfloor.$$ Then $$\mathcal W_{\sigma^2}(x,\;\cdot\;):=\iota_\ast\mathcal N_{x,\:\sigma^2}\;\;\;\text{for }x\in[0,1)$$ is a well-defined Markov kernel on $([0,1),\mathcal B([0,1)))$. Now observe that $$\mathcal W_{\sigma^2}(x,B)=\sum_{k\in\mathbb Z}\mathcal N_{x,\:\sigma^2}(k+B)=\sum_{k\in\mathbb Z}\mathcal N_{x-k,\:\sigma^2}(B)=\int_B\lambda({\rm d}y)\psi(y-x)$$ for all $(x,B)\in[0,1)\times\mathcal B([0,1))$.
Now define $$\mathcal M(x,\;\cdot\;):=p\mathcal U_{[0,\:1)}+(1-p)\mathcal W_{\sigma^2}(x_n,\;\cdot\;)\;\;\;\text{for }x\in[0,1)$$ and $$Q(x,\;\cdot\;):=\bigotimes_{n\in\mathbb N}\mathcal M(x_n,\;\cdot\;)\;\;\;\text{for }x\in[0,1)^{\mathbb N}.$$ Then, at a first glance, we should have $$Q(x,B)=\int_B\lambda({\rm d}y)q(y-x)\;\;\;\text{for all }(x,B)\in[0,1)^{\mathbb N}\times\mathcal B([0,1))^{\otimes\mathbb N}.\tag1$$ On the other hand, if $\sigma>\frac1{\sqrt{2\pi}}$, then $\varphi_{\sigma^2}<1$ and hence $q\equiv0$ ... So, something is wrong here. Does $Q$ admit a density (with respect to $\left.\lambda\right|_{[0,\:1)}^{\otimes\mathbb N}$) at all?
We've clearly got the following: If $I\subseteq\mathbb N$, let $$\pi_I:[0,1)^{\mathbb N}\to[0,1)^I\;,\;\;\;x\mapsto(x_i)_{i\in I}.$$ Now we should clearly have $$Q\left(x,\pi_I^{-1}\left(\times_{i\in I}B_i\right)\right)=\prod_{i\in I}\mathcal M(x_i,B_i)$$ for all finite $I\subseteq\mathbb N$ and $(B_i)_{i\in I}\subseteq[0,\infty)$.