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 1


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.

  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    Commented Jul 31, 2019 at 15:27
  • $\begingroup$ 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? $\endgroup$
    – 0xbadf00d
    Commented Jan 9, 2020 at 5:48
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jan 9, 2020 at 19:11
  • $\begingroup$ You're right, I missed that for a moment. $\endgroup$
    – 0xbadf00d
    Commented Jan 9, 2020 at 20:05

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