# Convergence of conditional distributions implies convergence of unconditional distributions

Let $$X,X_1,X_2,\dots$$ be real random variables defined on a probability space $$(\Omega,\mathcal A ,P)$$. Given a sub-$$\sigma$$-algebra $$\mathcal F\subset\mathcal A$$, let $$P_{X}$$ denote the distribution of $$X$$ and let $$P_{X|\mathcal F}$$ denote the conditional distribution of $$X$$ given $$\mathcal F$$, i.e. a Markov kernel from $$(\Omega,\mathcal F)$$ to $$(\mathbb R ,\mathcal B(\mathbb R))$$ such that $$\omega\mapsto P_{X|\mathcal F}(\omega,B)$$ is a version of $$P(X\in B|\mathcal F)$$ for each $$B\in \mathcal B(\mathbb R)$$. Suppose that

$$P_{X_n|\mathcal F}(\omega,\cdot)\Rightarrow P_{X|\mathcal F}(\omega,\cdot) \quad P\text{-almost surely},$$

where $$\Rightarrow$$ denotes convergence in distribution (weak convergence). (The above definition makes sense since conditional distributions are almost surely unique). Is it then also the case that $$P_{X_n}\Rightarrow P_X$$ ?

I tried the following:

Let $$f\in C_b(\mathbb R)$$. From the definition of weak convergence we have

$$\int f dP_{X_n|\mathcal F}(\omega,\cdot)\to \int f dP_{X|\mathcal F}(\omega,\cdot)\quad \text{as } n\to \infty, \quad P\text{-almost surely.}$$ Moreover, from the properties of conditional distributions we have

$$E[f\circ X|\mathcal F](\omega)=\int f dP_{X|\mathcal F}(\omega,\cdot) \quad P\text{-almost surely, }$$ for each $$n$$. Therefore $$E[f\circ X_n|\mathcal F]\to E[f\circ X|\mathcal F]$$ $$P$$-almost surely. From the dominated convergence theorem we get

$$E[f\circ X_n]=\int f dP_{X_n}\to \int f dP_X \quad \text{as } n\to \infty.$$

As $$f\in C_b(\mathbb R)$$ was arbitrary, we indeed have $$P_{X_n}\Rightarrow P_X$$.

Is this correct? Thanks a lot for your help.

This is a long comment that first better in the answer section:

Some attention is required to what versions of conditional probabilities are taken, for in general conditional probabilities are not representable as stochastic kernels unless some regularity assumptions are satisfied (which $$\mathbb{R}$$ satisfies).

Let $$X_\infty$$ denote $$X$$. Since $$\mathbb{R}$$ (with the usual Euclidean structure) is a nice space (Polish), for each $$n\in\mathbb{N}\cup\{\infty\}=\overline{\mathbb{N}}$$, the conditional probability $$P[X_n\in\cdot|\mathcal{F}]$$ is regular, that is there is a stochastic kernel $$\mu_n$$ from $$(\Omega,\mathcal{F}$$ to $$(\mathbb{R},\mathscr{B}(\mathbb{R})$$, and a null set $$N_n\in\mathcal{A}$$ such that

1. $$P[X_n\in B|\mathcal{F}](\omega)=\mu_n(\omega, B)$$ for all Borel set $$B$$ and $$\omega\in N_n$$
2. For any Borel set $$B$$, $$\omega\mapsto\mu_n(\omega,B)$$ is $$\mathcal{F}$$-measurable

A simple monotone class arguments shows that for any $$f\in\mathcal{C}_b(\mathbb{R})$$, \begin{align}E[f(X_n)|\mathcal{F}](\omega)=\int_\mathbb{R} f(x)\mu_n(\omega,dx),\qquad\omega\in N_n\tag{1}\label{one}\end{align}

For all $$\omega$$ outside $$N=\bigcup_{n\in\overline{\mathbb{N}}}N_n$$ (which is a negligible set) condition \eqref{one} holds.

If for any $$\omega\in\Omega\setminus N$$ we have that $$\mu_n(\omega,\cdot)$$ converges weakly to $$\mu_\infty(\omega,\cdot)$$, then by dominated convergence \begin{align} E[f(X_n)]&=\int_\Omega E[f(X_n)|\mathcal{F}](\omega)\,P(d\omega)\\ &=\int_{\Omega\setminus N}\Big(\int_\mathbb{R} f(x)\mu_n(\omega,dx)\Big)\,P(d\omega)\xrightarrow{n\rightarrow\infty}\int_{\Omega\setminus N}\Big(\int_\mathbb{R} f(x)\mu_\infty(\omega,dx)\Big)\,P(d\omega)\\ &=\int_\Omega E[f(X_\infty)|\mathcal{F}](\omega)\,P(d\omega)=E[f(X_\infty)] \end{align} Which is what you sketched in your posting.

The latter condition (on the kernels $$\mu_n)$$ however, seems to me that would be much harder to verify than to directly show that $$X_n$$ converges weakly in law.

• Thank you for your answer. I agree that verifying convergence of conditional distributions is much harder. Your calculations seem to confirm that my reasoning is correct: weak convergence of conditional distributions a.s. implies weak convergence in distribution? Jun 30 at 16:19
• Yes I was assuming real-valued random variables throughout. Jun 30 at 16:30
• I had a related question here : math.stackexchange.com/q/4188651/522332 . If you have any thoughts on this I would be grateful. Jul 2 at 13:56