# Can we show that $\mathbb R^{\mathbb N}\ni x\mapsto\bigotimes_{n\in\mathbb N}\mathcal N_{x,\:\sigma^2}$ is a Markov kernel?

Let $$\sigma>0$$ and $$\mathcal N_{x,\:\sigma^2}$$ denote the normal distribution with mean $$x\in\mathbb R$$ and variance $$\sigma^2$$. From the Ionescu-Tulcea theorem, we know that $$\kappa(x,\;\cdot\;):=\bigotimes_{n\in\mathbb N}\mathcal N_{x_n,\:\sigma^2}$$ is a well-defined probability measure on $$\mathcal B(\mathbb R)^{\otimes\mathbb N}$$ for all $$x\in\mathbb R^{\mathbb N}$$.

Are we able to show that $$\mathbb R^{\mathbb N}\ni x\mapsto\kappa(x,B)$$ is Borel measurable for all $$B\in\mathcal B(\mathbb R)^{\otimes\mathbb N}$$?

If the claim is true, $$\kappa$$ would be a Markov kernel on $$\left(\mathbb R^{\mathbb N},\mathcal B(\mathbb R)^{\otimes\mathbb N}\right)$$. Are we able to infer that then the restriction of $$\kappa$$ to $$[0,1]^{\mathbb N}\times\mathcal B([0,1])^{\otimes\mathbb N}$$ is a Markov kernel on $$\left([0,1]^{\mathbb N},\mathcal B([0,1])^{\otimes\mathbb N}\right)$$?

## 1 Answer

First of all, I suppose you mean $$\kappa$$ to be defined as $$\kappa(x,\;\cdot\;):=\bigotimes_{n\in\mathbb N}\mathcal N_{x_n,\:\sigma^2}$$ with $$x_n$$ on the right side instead of $$x$$, where $$x = (x_1, x_2, \dots)$$. As originally written it didn't make sense.

Defined thus, $$\kappa$$ is indeed a Markov kernel. As you note, we need to prove that $$\kappa(\cdot,B)$$ is measurable for every Borel set $$B\in\mathcal B(\mathbb R)^{\otimes\mathbb N}$$. One way is to use the Dynkin $$\pi$$-$$\lambda$$ lemma. Let $$\mathcal{L}$$ be the collection of all sets $$B\in\mathcal B(\mathbb R)^{\otimes\mathbb N}$$ such that $$\kappa(\cdot,B)$$ is Borel measurable. You may easily show that $$\mathcal{L}$$ is a $$\lambda$$-system:

• When $$B = \mathbb{R}^{\mathbb{N}}$$, we have $$\kappa(x,B)=1$$ for every $$x$$, and the constant function $$1$$ is measurable. So $$\mathbb{R}^{\mathbb{N}} \in \mathcal{L}$$.

• If $$B \in \mathcal{L}$$, then $$\kappa(x,B^c) = 1-\kappa(x,B)$$ for every $$x$$, because $$\kappa(x,\cdot)$$ is a probability measure. So $$\kappa(\cdot, B^c) = 1-\kappa(\cdot, B)$$ is measurable because $$\kappa(\cdot, B)$$ was.

• If $$B_1, B_2, \dots \in \mathcal{L}$$ are disjoint, and $$B = \bigcup_k B_k$$, then we have $$\kappa(\cdot, B) = \sum_{k=1}^\infty \kappa(\cdot, B_k)$$ which is measurable since it is an infinite sum of measurable functions.

Let $$\mathcal{P}$$ be the collection of all "rectangles" of the form $$B = B_1 \times B_2 \times \dots \times B_m \times \mathbb{R} \times \mathbb{R} \times \dots$$, where $$m$$ is an integer and $$B_1, \dots, B_m \in \mathcal{B}(\mathbb{R})$$. Clearly $$\sigma(\mathcal{P}) = \mathcal B(\mathbb R)^{\otimes\mathbb N}$$. So it remains to show that $$\mathcal{P} \subset \mathcal{L}$$. But for such $$B$$ we have $$\kappa(x,B) = \frac{1}{(2 \pi \sigma^2)^{m/2}} \prod_{k=1}^m \int_{B_k} \exp(-(x_k-y_k)^2/2\sigma^2)\,dy_k$$ and it is easy to check this is actually a continuous function of $$x$$.

You can also construct a proof with the monotone class theorem if you like it better. But either way, with some practice this kind of argument should be completely routine.

The Markov chain on $$\mathbb{R}^{\mathbb{N}}$$ whose transition kernel is $$\kappa$$ is easy to describe: the coordinates perform independent random walks, each of which has iid $$\mathcal{N}_{0, \sigma^2}$$ increments.

I am not sure I understand exactly what you mean by "restriction", but note for example that $$\kappa(x, [0,1]^{\mathbb{N}}) = 0$$ for all $$x$$. (If you sample countably many normal random variables all with the same variance, there is zero probability that they will all come out between 0 and 1.) So I don't see how to turn this into a Markov kernel on $$[0,1]^{\mathbb{N}}$$ in any sensible way.

• You're right, I've messed up the definition of $\kappa$. And I'm sorry, the "restriction to $[0,1]^{\mathbb N}\times\mathcal B([0,1])^{\otimes\mathbb N}$" doesn't make sense. – 0xbadf00d Jul 31 '19 at 15:27
• Does $\kappa$ admit a density $q$ with respect to $\lambda^{\otimes\mathbb N}$, where $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R)$? If $\varphi(x):=\frac1{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}$ for $x\in\mathbb R$, a candidate would be $q(x):=\prod_{n\in\mathbb N}\varphi(x_n)$ for $x\in(-1,1)^{\mathbb N}$. But this seems to be wrong, since if $\sigma>\frac1{\sqrt{2\pi}}$, then $\varphi<1$ and hence $q=0$. However, $q$ should be the correct density if we replace the infinite-product with a finite one, right? – 0xbadf00d Jan 9 '20 at 5:48
• @0xbadf00d: There is no such measure as $\lambda^{\otimes \mathbb{N}}$ on $\mathbb{R}^{\mathbb{N}}$. The usual construction of infinite product measure only works for probability measures. – Nate Eldredge Jan 9 '20 at 19:11
• You're right, I missed that for a moment. – 0xbadf00d Jan 9 '20 at 20:05