# Construction of a coupling of a sequence of Bernoulli random variables

Let $$(X_n)_{n\in\mathbb{N}}$$ be a sequence of Bernoulli distributed random variables defined on a probability space $$(\Omega,\mathcal{F},P)$$ and $$(\mathcal{F}_n)_{n\in\mathbb{N}}$$ be a filtration such that, for all $$n$$, $$X_n$$ is measurable with respect to $$\mathcal{F}_{n+1}$$ and $$E\lbrack X_n|\mathcal{F}_n\rbrack>1-\delta$$ for some fixed $$\delta\in(0,1)$$. I want to construct a probability space $$(\mathcal{X},\mathcal{A},\mu)$$ and sequences $$(Y_n)_{n\in\mathbb{N}}$$ and $$(Z_n)_{n\in\mathbb{N}}$$ of random variables which are defined on this probability space such that $$(Z_n)_{n\in\mathbb{N}}\overset{d}{=}(X_n)_{n\in\mathbb{N}}$$, $$Y_1,Y_2,\ldots$$ are iid Bernoulli distributed random variables with parameter $$1-\delta$$ and $$Z_n\geq Y_n$$ for all $$n$$. Note that by the assumptions above, we can conclude that, for all $$n\in\mathbb{N}$$ and $$i\in\{0,1\}^n$$, we have $$P(X_1=i_1,\ldots,X_n=i_n,X_{n+1}=1)\geq (1-\delta)P(X_1=1,\ldots,X_n=i_n).$$ First I will give you the basic idea that I have: My idea is to define $$Y_n$$ by using iid random variables $$(U_n)_{n\in\mathbb{N}}$$ which are uniformly distributed on $$(0,1)$$ and setting $$Y_n=1_{\{U_n<1-\delta\}}$$ for all $$n$$. Now I want to define $$(Z_n)_{n\in\mathbb{N}}$$ recursively: On the set $$\{Y_1=1\}$$ we set $$Z_1=1$$. I want $$Z_1$$ to have the same distribution as $$X_1$$, so I must define $$Z_1$$ on the set $$\{Y_1=0\}$$ so that $$\mu(Y_1=0,Z_1=1)=P(X_1=1)-(1-\delta)$$. At this point I am not sure how to do this. I was thinking about using a random variable $$V_1$$, which is also uniformly distributed on $$(0,1)$$ and independent of $$(U_n)_{n\in\mathbb{N}}$$ and setting \begin{align*}Z_1=1_{\{Y_1=1\}}+1_{\{Y_1=0,V_1<(P(X_1=1)-(1-\delta))/\delta\}}. \end{align*} Then $$\mu(Z_1=1)=P(X_1=1)$$, is this construction correct? Next, I would construct $$Z_2$$. My idea is to use the same idea as in the construction of $$Z_1:$$ Let $$V_2,V_3$$ be two random variables which are uniformly distributed on $$(0,1)$$, independent of each other, independent of $$V_1$$ and independent of $$(U_n)_{n\in\mathbb{N}}$$. Then I would set \begin{align*} Z_2=1_{\{Y_2=1,Z_1=0\}}+1_{\{Y_2=1,Z_1=1\}}+1_{\{Y_2=0,Z_1=0,V_2<(P(X_1=0,X_2=1)-(1-\delta)P(X_1=0))/(\delta P(X_1=0))\}}+1_{\{Y_2=0,Z_1=1,V_3<(P(X_1=1,X_2=1)-(1-\delta)P(X_1=1))/(\delta P(X_1=1))\}}. \end{align*} By this construction, $$(Z_1,Z_2)$$ should have the same distribution as $$(X_1,X_2)$$ if I did not make a mistake. Analogously, I would now define $$Z_3,Z_4$$ and so on. Ultimately, this leads to a process $$(Z_n)_{n\in\mathbb{N}}$$ such that for all $$n$$ we have $$(Z_1,\ldots,Z_n)\overset{d}{=}(X_1,\ldots,X_n)$$, i.e. $$(Z_n)_{n\in\mathbb{N}}\overset{d}{=}(X_n)_{n\in\mathbb{N}}$$ by Kolmogorov's extension theorem. The problem that I have is that I do not know how to make this proof rigorously. In the $$n$$-th step of the construction we need $$2^{n-1}+1$$ uniformly distributed random variables: one for the construction of $$Y_n$$ and $$2^{n-1}$$ more to complete the construction of $$Z_n$$, i.e. I would define $$(\Omega_{n,k},\mathcal{F}_{n,k},P_{n,k}):=((0,1),\mathcal{B}(0,1),\text{Uni}(0,1))$$ for all $$n\in\mathbb{N}$$ and $$k=0,\ldots,2^{n-1}$$ and set \begin{align*} (\mathcal{X},\mathcal{A},\mu):=\left(\times_{n\in\mathbb{N}}\left(\times_{k=0}^{2^{n-1}}\Omega_{n,k}\right),\otimes_{n\in\mathbb{N}}\left(\otimes_{k=0}^{2^{n-1}}\mathcal{F}_{n,k}\right),\otimes_{n\in\mathbb{N}}\left(\otimes_{k=0}^{2^{n-1}}P_{n,k}\right)\right). \end{align*} Now I am not sure how to make the construction of the processes rigorously. If we assume that we have already constructed $$Y_1,\ldots,Y_n$$ and $$Z_1,\ldots,Z_n$$, I would define \begin{align*} Y_{n+1}(x)=1_{\{x_{n+1,0}<1-\delta\}}(x), \end{align*} where for $$x\in\mathcal{X}$$, $$x_{m,j}$$ denotes the coordinate of $$x$$ which belongs to $$\Omega_{m,j}$$. But now I am not sure how to formally construct $$Z_{n+1}$$. Does anyone have an idea?

Construction.

• Let $$(U_n)_{n\in\mathbb{N}}$$ be a sequence of i.i.d. random variables which are uniformly distributed on $$(0, 1)$$.

• Then, define $$p_n : \{0, 1\}^{n-1} \to [0, 1]$$ by \begin{align*} p_n(x_{[1:n)}) &= \mathbf{P}(X_n = 1 \mid X_{[1:n)} = x_{[1:n)}) \\ &= \mathbf{E}[X_n \mid X_{[1:n)}=x_{[1:n)}], \end{align*} where $$x_{[1:n)} = (x_1, \ldots, x_{n-1})$$ and likewise for $$X_{[1:n)}$$. (Here, $$x_{[1:1)}$$ is regarded as an empty list $$\varnothing$$, so that $$p_1(\varnothing) = \mathbf{P}(X_1 = 1) = \mathbf{E}[X_1]$$.)

• Finally, define $$Y_n$$ and $$Z_n$$ by \begin{align*} Y_n = \mathbf{1}_{\{U_n \leq 1-\delta\}} \qquad\text{and}\qquad Z_n = \mathbf{1}_{\{ U_n \leq p_n(Z_{[1:n)}) \}}. \end{align*}

Observations. From the above construction, we observe:

1. $$(Y_n)_{n\in\mathbb{N}}$$ is a sequence of i.i.d. $$\text{Bernoulli}(1-\delta)$$-random variables.

2. $$Z_n \geq Y_n$$ holds, since $$p_n(x_{[1:n)}) > 1 - \delta$$.

3. For any $$n \in \mathbb{N}$$, note that $$Z_{[1:n)}$$ is a deterministic function of $$U_{[1:n)}$$. In particular, $$Z_{[1:n)}$$ is independent of $$U_n$$. Then, for any $$z_{[1:n)} \in \{0, 1\}^{n-1}$$, we have \begin{align*} \mathbf{P}(Z_n = 1 \mid Z_{[1:n)} = z_{[1:n)}) &= \mathbf{P}(U_n \leq p_n(Z_{[1:n)}) \mid Z_{[1:n)} = z_{[1:n)}) \\ &= p_n(z_{[1:n)}) \\ &= \mathbf{P}(X_n = 1 \mid X_{[1:n)} = z_{[1:n)}). \end{align*} So by the principle of mathematical induction, $$(Z_1, \ldots, Z_n) \stackrel{d}= (X_1, \ldots, X_n)$$ holds for all $$n \in \mathbb{N}$$, which in turn implies that $$(Z_n)_{n\in\mathbb{N}} \stackrel{d}= (X_n)_{n\in\mathbb{N}}$$.

Conclusion. The sequences $$(Z_n)_{n\in\mathbb{N}}$$ and $$(Y_n)_{n\in\mathbb{N}}$$ constructed as above satisfies all the desired properties.