Distribution of $[0,1]$-valued random variable determined by distribution of $\{0,1\}$-valued exchangeable sequence

Question

I'm trying to use de Finetti's theorem to show that if $Y$ is a random variable with values in $[0,1]$, then the distribution of $Y$ is uniquely determined by its moments $\mathbb E[Y^n] =: m_n$. This can be proven using separating families of functions (things like characteristic functions, moment-generating functions, etc), but I’d like to use a different strategy to build my intuition about exchangeability and conditional expectation.

I've shown previously the following:

Suppose $X = (X_n)_{n \in \mathbb Z}$ is an exchangeable sequence of random variables and $X_n \in \{0,1\}$ for all $n$. Then the distribution of $X$ is uniquely determined by the values $\mathbb E[X_1 X_2 \cdots X_n] =: m_n$.

So, let $X = (X_n)_{n \in \mathbb N}$ be a sequence of random variables with values in $\{0,1\}$ for which, given $Y$, the sequence is independent and $\mathrm{Ber}_Y$-distributed. That is, for any finite $J \subset \mathbb Z$, and $k_j \in \{0,1\}$ for $j \in J$, $$ \mathbb P\left[ \mathop\bigcap_{j \in J} \{X_j = k_j\} \: \middle|\: Y \right] = \prod_{j \in J} \mathbb P\left[ X_j = k_j\:\middle|\: Y\right] \quad \textrm{and} \quad \mathbb P[X_n = 1\:|\:Y] = 1-\mathbb P[X_n = 0 \:|\: Y] = Y. $$ (Intuitively, we can think of $Y$ as the random unfair weight of a coin, and $(X_n)$ as the result of flipping the unfairly weighted coin.)

It turns out that $\mathbb E[Y^n] = \mathbb E[X_1 \cdots X_n]$, which follows from a calculation with regular conditional distributions. And, by de Finetti’s theorem, $X$ is exchangeable. So, using the above result, the distribution $\mathcal L[X]$ is uniquely determined by the moments of $Y$.

My question: If $Y$ is a $[0,1]$-valued random variable, and $X = (X_n)_{n \in \mathbb Z}$ is independent given $Y$ and $\mathrm{Ber}_Y$-distributed given $Y$ (to put it another way, $X$ is $\mathrm{Ber}_Y^{\otimes \mathbb Z}$-distributed given $Y$), is the distribution of $Y$ determined uniquely by the distribution of $X$?

De Finetti's theorem also tells us that there is a random variable $Z$ with values in $[0,1]$ for which, given $Z$, the sequence $(X_n)$ is independent and $\mathrm{Ber}_Z$-distributed. The issue is this random variable (and its distribution) need not be uniquely defined by that theorem, at least the version I'm familiar with. However, using regular conditional distributions, we note that for $A \subset \{0,1\}^{\mathbb Z}$ measurable: $$ \mathbb P[X \in A] = \int_{[0,1]} \mathbb P[X \in A \: | \: Y = y] \, \mathbb P_Y[dy] = \int_{[0,1]} \mathrm{Ber}_y^{\otimes \mathbb Z}(A) \, \mathbb P_Y[dy]. $$ In particular, $\int_{[0,1]} \mathrm{Ber}_y^{\otimes \mathbb Z}(A) \, \mathbb P_Y[dy] = \int_{[0,1]} \mathrm{Ber}_y^{\otimes \mathbb Z}(A) \, \mathbb P_Z[dy]$ for all measurable $A \subset \{0,1\}^{\mathbb Z}$. But concluding that the distribution of $Y$ is uniquely determined from the functions $\mathrm{Ber}_y^{\otimes \mathbb Z}(A)$ is basically using separating classes of functions. I feel like there has to be a more natural way.

Any thoughts?

Answer 1

The condition follows from the following two lemmas:

Lemma 1. If $X = (X_n)_{n \in \mathbb N}$ is an exchangeable process with values in $\{0,1\}$, then the distribution $\mathcal L[X]$ is uniquely determined by the values $m_n := \mathbb E[X_1 \cdots X_n]$.

Proof. This is part (i) of the exercise. QED.

Lemma 2 (Conditional Strong Law of Large Numbers). If $\mathcal F \subset \mathcal A$ is a $\sigma$-algebra and $X = (X_n)_{n \in \mathbb N}$ is a square-integrable stochastic process for which, given $\mathcal F$, we have that $X$ is independent and identically distributed, then: $$ \mathbb P\left[\limsup_{n \to \infty} \left| \frac 1 n \sum_{k=1}^n \left(X_k - \mathbb E\left[X_k\middle|\mathcal F\right]\right)\right| = 0\right]= 1. $$

Proof. This follows from the conditional Chebyshev inequality and the conditional Borel-Cantelli lemma, as well as the argument to show the un-conditional version of this lemma found in, for example, Theorem 5.16 in Achim Klenke's "Probability Theory: A Concrete Introduction", ed. 3. QED.

Let $Y$ be a random variable on $[0,1]$, and let $X = (X_n)_{n \in \mathbb N}$ be a sequence of random variables in $\{0,1\}$ for which, given $Y$, the sequence $X$ is independent and $\mathrm{Ber}_Y$-distributed. By de Finetti's Theorem, $X$ is exchangeable. And, for any $n \in \mathbb N$, $$ \mathbb E\left[ X_1 \cdots X_n\right] = \mathbb E\left[ \mathbb E\left[X_1 \cdots X_n \middle| Y\right]\right] = \mathbb E\left[ \mathbb E\left[X_1\middle| Y\right] \cdots \mathbb E\left[X_n\middle| Y\right]\right] = \mathbb E\left[Y^n\right] = m_n. $$ So, the moments of $Y$ uniquely determine the distribution $\mathcal L[X]$ by Lemma 1.

Let $F : \{0,1\}^\mathbb N \to [0,1]$ be given by $$ F(x) = F\left(\left(x_n\right)_{n \in \mathbb N}\right) = \limsup_{n \to \infty} \frac 1 n \sum_{k=1}^n x_n = \limsup_{n \to \infty} \frac 1 n \# \left\{ k \in \{1, \ldots, n\} : x_k = 1\right\} . $$ We claim $Y = F(X)$ almost surely. Indeed, since the $X_n$ take values in $\{0,1\}$, and since $\mathbb E[X_k|Y] = Y$ for all $k \in \mathbb N$, by the conditional strong law of large numbers (Lemma 2 above), \begin{align*} \mathbb P\left[ F(X) \neq Y\right] &= \mathbb P\left[ \limsup_{n \to \infty} \frac 1 n \sum_{k=1}^n \left(X_k - Y\right) \neq 0\right]= 0. \end{align*} Therefore, if $A \subset [0,1]$, $$ \mathbb P[Y \in A] = \mathbb P[F(X) \in A] = \mathbb P[X \in F^{-1}(A)] = \mathcal L[X]\left(F^{-1}(A)\right). $$ Since $\mathcal L[X]$ is uniquely determined by the moments of $Y$, this shows that $\mathbb P_Y$ is also determined uniquely by the moments of $Y$. QED.

Answer 2

Let $1_{(a,b)}(y)$ be indicator function of the open interval $(a,b)$.

There exists a sequence of continuous functions $g_{\epsilon}(y)$ with $|g_{\epsilon}(y)| \leq 1$ such that as $\epsilon \rightarrow 0$ we have $g_{\epsilon}(y) \rightarrow 1_{(a,b)}(y)$ in sup norm over $y \in [0,1] \setminus ([a-\epsilon,a+\epsilon] \cup [b - \epsilon, b+\epsilon])$.

Now let $p_{n,\epsilon}(y)$ be a sequence of polynomials such that as $n \rightarrow \infty$, $p_{n,\epsilon}(y) \rightarrow g_{\epsilon}(y)$ in sup norm over $y \in [0,1]$.

We know the value of $\int_{0}^1 p_{n,\epsilon}(y) f_Y(y) dy$

Now $\lim_{n\rightarrow \infty, \epsilon \rightarrow 0}\int_{0}^1 p_{n,\epsilon}(y) f_Y(y) dy = \int_{0}^1 1_{(a,b)}(y) f_Y(y) dy = P(Y \in (a,b))$. where we assumed that $f_Y(y)$ is continuous at $y = a,b$.

Hence in the limit we know the distribution of $Y$ assuming $f_Y(y)$ is continuous.

We are using the fact that polynomials are dense in $C([0,1])$. This amounts to a separating family of functions. But i feel the above is a slightly different perspective compared to characteristic functions. The question is interesting.