Independence

Let $(\Omega, \mathcal{F}, \Pr)$ be a probability space.

Independence Events

Definition. A finite collection of events $A_1, \ldots, A_n \in \mathcal{F}$ are independent, if for each collection of events $B_1, \ldots, B_n$, for which either $B_i = \Omega$ or $B_i = A_i$, $i \in [n]$, it holds that

$$\Pr[ B_1 \cap \ldots \cap B_n ] = \Pr[ B_{1} ] \cdots \Pr[ B_{n} ].$$

Since every $B_i = \Omega$ can be suppressed on each side of the equation, this is equivalent to that for each set of distinct indices $i_1 ,i_2 \ldots ,i_k$ between $1$ to $n$,

Definition. An infinite (perhaps uncountable) collection of events $\{ A_\theta, \theta \in \Theta \}$ are independent if each of its finite sub-collections of events are independent.

Independence Classes

Definition. A finite classes $\mathcal{A}_1, \ldots, \mathcal{A}_n \subseteq \mathcal{F}$ are independent, if for each choice of $A_i \in \mathcal{A}_i, i \in [n]$, the events $A_i, i \in [n]$ are independent.

Definition. An infinite (perhaps uncountable) collection of classes $\{ \mathcal{A}_\theta : \theta \in \Theta \}$ are independent if for each choice of $A_\theta \in \mathcal{A}_\theta, \theta \in \Theta$, the events $\{ A_\theta, \theta \in \Theta \}$ are independent.

Observe that by the definition of independence of infinite collection of event, the independence of $\{ \mathcal{A}_\theta : \theta \in \Theta \}$ is equivalent to the independence of each finite sub-collection $\mathcal{A}_{\theta_1}, \ldots, \mathcal{A}_{\theta_k}$ of the class.

Independent $\sigma$-fields

The following theorem underpins a number of theorems related to independence of random variables. It states that to decide a collection of $\sigma$-fields are independent, it suffices to show that a collection of $\pi$-systems generating them are independent.

Theorem. If the classes $\mathcal{A}_1, \ldots, \mathcal{A}_n \subseteq \mathcal{F}$ are independent and $\pi$ systems, then $\sigma(\mathcal{A}_1), \ldots, \sigma(\mathcal{A}_n)$ are independent.

Proof. Fix $A_2 \in \mathcal{A}_2, \ldots, A_n \in \mathcal{A}_n$.

Define

Lemma. $\mathcal{L}$ is a $\lambda$ system.

Since $\mathcal{L}$ contains the $\pi$ system $\mathcal{A}_1$, by $\pi$-$\lambda$ theorem, it follows that $\sigma(\mathcal{A}_1) \subseteq \mathcal{L}$. Therefore, $\sigma(\mathcal{A}_1), \ldots, \mathcal{A}_n$ are independent.

Repeating the process proves that $\sigma(\mathcal{A}_1), \sigma(\mathcal{A}_2), \ldots, \mathcal{A}_n$ are independent, and so on.

Proof of the Lemma.

1. $\Omega \in \mathcal{L}$.

2. If $E \in \mathcal{F}$, then $\bar E \in \mathcal{F}$: if

   $$\Pr[ E \cap B_{2} \cap \ldots \cap B_{k} ] = \Pr[E] \cdot \Pr[ B_2 ] \cdots \Pr[ B_n ].$$

   where $B_i = A_i$ or $B_i = \Omega$, $i \in [2 .\,.\, n]$, then

   $$\begin{aligned} \Pr[ \bar E \cap B_2 \cap \ldots \cap B_n ] &= \Pr[ B_2 \cap \ldots \cap B_n ] - \Pr[ E \cap B_2 \cap \ldots \cap B_n ] \\ &= (1 - \Pr[E]) \cdot \Pr[ B_2 ] \cdots \Pr[ B_n ] \\ &= \Pr[\bar E] \cdot \Pr[ B_2 ] \cdots \Pr[ B_n ] \end{aligned}$$

3. If $E_k \in \mathcal{F}, k \in \mathbb{N}^+$ are pairwise disjoint, and $E = \cup_{k \ge 1} E_k$, then $E \in \mathcal{F}$:

   $$\begin{aligned} \Pr[ E \cap B_2 \cap \ldots \cap B_n ] &= \sum_k \, \Pr[ E_k \cap B_2 \cap \ldots \cap B_n ] \\ &= \sum_k \, \Pr[E_k] \cdot \Pr[ B_2 ] \cdots \Pr[ B_n ] \\ &= \Pr[E] \cdot \Pr[ B_2 ] \cdots \Pr[ B_n ]. \end{aligned}$$

$\square$

Simple Random Variable

Definition. A random variable $X : \Omega \rightarrow \mathbb{R}$ is simple, if it has finite range (assumes finitely many values), and for each $x \in \mathbb{R}$,

$$\{ \omega \in \Omega : X(\omega) = x \} \in \mathcal{F}.$$

In probability theory, $\omega$ is often omitted, and $\{X = x \}$ is short for $\{ \omega \in \Omega : X(\omega) = x \}$. It it easy to see that a random variable can be represented by

where the $A_i \in \mathcal{F}, i \in [n]$ is a finite partition of $\Omega$.

Definition. For a finite or countably infinite sequence $X_1, X_2, \ldots$ of random variables, $\sigma(X_1, X_2, \ldots)$ is the intersection of all $\sigma$-fields with respect to which each $X_i$ is measurable.

Independence

Definition. A finite or countably infinite sequence $X_1, X_2, \ldots$ of simple random variables are independent, if $\sigma(X_1), \sigma(X_2), \ldots$ are independent.

We study a several importance properties.

Property 1

Theorem. A finite sequence $X_1, \ldots, X_n$ of simple random variables are independent, if and only if for each $x_1, \ldots, x_n \in \mathbb{R}$,

$$\Pr[ X_1 = x_1, \ldots, X_n = x_n ] = \Pr[ X_1 = x_1 ] \cdots \Pr[ X_n = x_n ].$$

Proof. If $X_1, \ldots, X_n$ of random variables are independent, the equation clearly holds. It suffices to prove the other direction.

For each $i \in [n]$, let $X_i(\Omega)$ be the range of $X_i$, which is finite. Define the class

which is a $\pi$-system, since the $X_i^{-1} (x), x \in X_i(\Omega)$ are pairwise disjoint. Then $\mathcal{A}_1, \ldots, \mathcal{A}_n$ are independent. It follows that $\sigma(X_1), \sigma(X_2), \ldots$ are independent.

$\square$

Property 2

Next, we prove an important result. Consider a finite sequence of simple random variables:

For each $i \in [n]$, $f_i: \mathbb{R}^{k_i} \rightarrow \mathbb{R}$ is a function. Then

Claim. The $f_i(X_{i, 1}, \ldots X_{k_i})$ are independent.

The following theorem states that functions of finite simple random variables are again simple random variables.

Theorem. For a finite sequence $X_1, \ldots, X_n$ of simple random variables,

1. The $\sigma$-field $\sigma(X_1, \ldots, X_n)$ consists of the sets

   $$\{ \omega \in \Omega : \left( X_1(\omega), \ldots, X_n(\omega) \right) \in H \},$$

   where $H \in \mathcal{R}^n$; $H$ in this representation can be taken finite.

2. A simple random variable $Y$ is measurable $\sigma(X_1, \ldots, X_n)$ if and only if

   $$Y = f(X_1, \ldots, X_n),$$

   for some $f: \mathbb{R}^n \rightarrow \mathbb{R}$.

Proof of Claim One. Define the function $\pmb{X}: \Omega \rightarrow \mathbb{R}^n$ by

For each $H \in \mathcal{R}^n$, define $\pmb{X}^{-1} (H) \doteq \{ \omega \in \Omega : \pmb{X} (\omega) \in H \}$, and define and the class of sets

Step One: $\pmb{X}^{-1} (\mathcal{R}^n) \subseteq \sigma(X_1, \ldots, X_n)$.

Observe that for each $\pmb{x} = (x_1, \ldots, x_n) \in \mathbb{R}^n$,

Since $\{ \omega \in \Omega : X_i(\omega) = x_i \} \in \sigma(X_1, \ldots, X_n)$ for each $i \in [n]$, we have

Since $\pmb{X}$, as a function, has finite range, each set $ \pmb{X}^{-1} (H) $ is a finite union of sets of the form $\{ \omega \in \Omega : \pmb{X} (\omega) = \pmb{x} \}$, and therefore belongs to $\sigma(X_1, \ldots, X_n)$.

Step One: $\sigma(X_1, \ldots, X_n) \subseteq \pmb{X}^{-1} (\mathcal{R}^n)$.

It suffices to prove that $\pmb{X}^{-1} (\mathcal{R}^n)$ a $\sigma$-field. At a high level, this follows from that $\mathcal{R}^n$ is a $\sigma$-field.

1. Since if $H \in \mathcal{R}^n$, then $\bar H \in \mathcal{R}^n$, it holds that

   $$\overline{\pmb{X}^{-1} (H)} = \pmb{X}^{-1} (\bar H) \in \pmb{X}^{-1} (\mathcal{R}^n).$$

2. Since if $H_k \in \mathcal{R}^n, k \in \mathbb{N}^+$, then $\cup_{k \ge 1} H_k \in \mathcal{R}^n$, it holds that

   $$\cup_{k \ge 1} \pmb{X}^{-1} (H_k) = \pmb{X}^{-1} (\cup_{k \ge 1} H_k) \in \pmb{X}^{-1} (\mathcal{R}^n).$$

$\square$

Proof of Claim Two. Suppose that

for some $f: \mathbb{R}^n \rightarrow \mathbb{R}$. Since $\pmb{X}$ has finite range, so does $Y$. For each possible output $y \in \mathbb{R}$, $Y^{-1} (y)$ is a finite union of $\pmb{X}^{-1}(\pmb{x})$ over those $\pmb x$ for which $f(\pmb{x}) = y$. It follows that $Y$ is measurable $\sigma(X_1, \ldots, X_n)$.

To prove the other direction, suppose that $Y$ is measurable $\sigma(X_1, \ldots, X_n)$. Since it is a simple random variable, it assumes finite values $y_1, \ldots, y_k$.

• The $\sigma(X_1, \ldots, X_n)$-sets $Y^{-1} (y_i), i \in [k]$ forms a partition of $\Omega$.

• Since $Y^{-1} (y_i) \in \sigma(X_1, \ldots, X_n)$, via previous claim, there exists for each $i \in [n]$, a set $H_i \in \mathcal{R}^n$, such that $\pmb{X}^{-1}(H_i) = Y^{-1} (y_i)$ and $H_i$ can be taken finite.

• For each $H_i$, we construct a subset $H_i' \subseteq H_i$, by removing from $H_i$ the $\pmb{x}$ for which $\pmb{X}^{-1} (\pmb{x}) = \emptyset$. The $H_i', i \in [n]$ are pairwise disjoint.

• Finally, for each $i \in [n]$ and each $\pmb{x} \in H_i'$, define $f(\pmb{x}) = y_i$. For all other $\pmb{x} \notin \cup_{i \in [n]} H_i'$, define $f(\pmb{x}) = 0$.

$\square$

We finish the proof of the claim with another result.

Theorem. Let

$$\begin{matrix} X_{1, 1}, &X_{1, 2}, &\ldots \\ X_{2, 1}, &X_{2, 2}, &\ldots \\ \ldots &\ldots &\ldots \end{matrix}$$

be a (possibly countably infinite) sequence of independent simple random variables. For each $n \in \mathbb{N}^+$, suppose the random variable $Y_n$ is measurable $\mathcal{F}_n \doteq \sigma(X_{n, 1}, X_{n, 2}, \ldots )$. Then $Y_1, Y_2, \ldots$ are independent.

Proof. For each $n \in \mathbb{N}^+$, define the class

which is the finite intersections of the sets $X_{n, k}^{-1} (x), k \in \mathbb{N}^+, x \in \mathbb{R}$. It is easy to see that $\emptyset \in \mathcal{A}_n$, and $\mathcal{A}_n$ is a $\pi$-system. Further,

• $\mathcal{A}_n$ generates $\mathcal{F}_n$, i.e., $\sigma(\mathcal{A}_n) = \mathcal{F}_n$:

  • $\sigma(\mathcal{A}_n) \subseteq \mathcal{F}_n$: via definition of $\mathcal{F}_n$, $X_{n, k}^{-1} (x) \in \mathcal{F}_n$, for each $k \in \mathbb{N}^+, x \in \mathbb{R}$. Hence, each element in $\mathcal{A}_n$ belongs to $\mathcal{F}_n$, and $\sigma(\mathcal{A}_n) \subseteq \mathcal{F}_n$.

  • $\sigma(\mathcal{A}_n) \supseteq \mathcal{F}_n$: via definition of $\mathcal{A}_n$, $X_{n, k}^{-1} (x) \in \mathcal{A}_n$, for each $k \in \mathbb{N}^+, x \in \mathbb{R}$. Hence, each $X_{n, k}$ is measurable $\sigma(\mathcal{A}_n)$. Via definition of $\mathcal{F}_n$, $\sigma(\mathcal{A}_n) \supseteq \mathcal{F}_n$.

• The $\mathcal{A}_n$ are independent: let $n_1, \ldots, n_m$ be a finite sequence of indexes in $\mathbb{N}^+$. Suppose that for each $j \in [m]$,

  $$B_{n_j} \doteq X_{n_j, i_1(j)}^{-1} (x_{i_1(j)}) \cap \ldots \cap X_{n_j, i_k(j)}^{-1} (x_{i_{k}(j)}).$$

  Then

  $$\begin{aligned} \Pr[ B_{n_1} \cap \ldots \cap B_{n_m} ] &= \Pr\left[ \cap_{j \in [m]} \left( X_{n_j, i_1(j)}^{-1} (x_{i_1(j)}) \cap \ldots \cap X_{n_j, i_k(j)}^{-1} (x_{i_{k}(j)}) \right) \right] \\ &= \prod_{j \in [m]} \Pr[ X_{n_j, i_1(j)} = x_{i_1(j)}] \cdots \Pr[X_{n_j, i_k(j)} = x_{i_{k}(j)} ] \\ &= \prod_{j \in [m]} \Pr[B_{n_j}]. \end{aligned}$$

It follows that the $\mathcal{F}_n$ are independent.

$\square$

Remark. The second claim would be much easier to prove, if $f$ is measurable $\mathcal{R}^n / \mathcal{R}$.

Random Variables and Vectors

Definition. A function $X : \Omega \rightarrow \mathbb{R}$ is a random variable, if it is measurable $\mathcal{F} / \mathcal{R}$ (or measurable $\mathcal{F}$ for short), i.e.,

$$[X \in H] = X^{-1}(H) \in \mathcal{F}, \forall H \in \mathcal{R}.$$

Definition. A function $X : \Omega \rightarrow \mathbb{R}^k$ is a random vector, if it is measurable $\mathcal{F} / \mathcal{R}^k$ (or measurable $\mathcal{F}$ for short), i.e.,

$$[X \in H] = X^{-1}(H) \in \mathcal{F}, \forall H \in \mathcal{R}^k.$$

Consider a function $X : \Omega \rightarrow \mathbb{R}^k$. For each $\omega \in \Omega$, we write the $i$-th coordinate of $X(\omega)$ as $X_i(\omega)$. Hence, $X(\omega) = (X_1(\omega), \ldots, X_k(\omega))$, and each $X_i$ is itself a function from $\Omega$ to $\mathbb{R}$.

Theorem. $X$ is measurable $\mathcal{F} / \mathcal{R}^k$ (a random vector), if and only if each $X_i$ is measurable $\mathcal{F} / \mathcal{R}$ ( a random variable).

The proof relies on the following lemma.

Lemma. $\mathcal{L} \doteq \{ H \in \mathcal{R}^k : [X \in H] \in \mathcal{F} \} \subseteq \mathcal{R}^k$ is a $\sigma$-field.

Proof of the Lemma. Observe that $\mathcal{L}$ is closed under complement and countable union:

$\square$

Proof of the Theorem.

Define

for each $H \in \mathcal{S}$,

$\impliedby:$ Suppose that each $X_i$ is measurable, then $[X_i \le x_i] \in \mathcal{F}, i \in [k]$. It follows that $[X \in H] \in \mathcal{F}$, and $\mathcal{S} \subseteq \mathcal{L}$. Since $\mathcal{S}$ generates $\mathcal{R}^k$, we have $\mathcal{L} = \mathcal{R}^k$.

$\implies:$ Suppose that $X$ is measurable. Fix an $i \in [k]$. By setting $x_j = \infty$ for each $j \neq i$ we have

It concludes that $X_i$ is measurable.

$\square$

Theorem. The $\sigma$-field $\sigma(X) = \sigma(X_1, \ldots, X_k)$ consists exactly of the sets $[X \in H]$ for $H \in \mathcal{R}^k$.

Proof. Let $\mathcal{G} \doteq \{ [X \in H] : H \in \mathcal{R}^k \} \subseteq \mathcal{F}$. Then $\mathcal{G}$ is closed under complement and countable union:

Hence $\mathcal{G}$ is a $\sigma$-field, and $\sigma(X) \subseteq \mathcal{G}$. But by the definition of $\sigma(X)$ and $\mathcal{G}$, we also have $\mathcal{G} \subseteq \sigma(X)$.

$\square$

Theorem. In order that a random variable $Y$ be measurable $\sigma(X_1, \ldots, X_k)$, it is necessary and sufficient that there exists a measurable map $f: \mathbb{R}^k \rightarrow \mathbb{R}$ s.t., $Y(\omega) = f(X_1(\omega), \ldots, X_k(\omega))$ for all $\omega$.

Proof.

$\impliedby:$ this direction is obvious.

$\implies:$ first, assume that $Y$ is a simple random variable which assumes value $y_1, \ldots, y_n$. Then the $\mathcal{F}$-sets $[Y = y_i], i \in [n]$ constitute a partition of $\Omega$. Via the previous theorem, there exists $H_1, \ldots, H_n \in \mathcal{R}^k$, such that $[Y = y_i] = [X \in H_i]$.

Note that the $H_i$ are not necessarily disjoint. But we have the following observation: fix an $[Y = y_i]$. Then

The second observation is true, as otherwise, $X^{-1}(H_j) = [Y = y_j]$ has non-empty intersection with $[Y = y_i]$, a contradiction. At this point, we can set

and set $\bar H_{n + 1} = \mathcal{R}^k - \cup_{i \in [n]} \bar H_i$. Then it still holds that $[X \in \bar H_i] = [Y = y_i], \forall i \in [n]$. Define the function $f: \mathbb{R}^k \rightarrow \mathbb{R}$ by

Clearly it holds that $Y = f \circ X$.

Next, we treat the general case. There exists a sequence of simple random variables $Y_n \rightarrow Y$. Let $f_n$ be the corresponding simple functions for which $Y_n = f_n \circ X$. Since the $f_n$ are measurable, so are the functions $\sup_n f_n$ and $\inf_n f_n$. It follows that the set $C \doteq \{ t \in \mathbb{R}^k : \sup_n f_n(t) = \inf_n f_n(t) \}$ for which the $f_n$ converges lies in $\mathcal{R}^k$. Define the function $f$ by

It is easy to check that $f$ is measurable. For each $\omega \in \Omega$, we must have $Y(\omega) = \lim_n Y_n (\omega) = \lim_n f_n(X(\omega))$; hence $X(\omega) \in C$, and $\lim_n f_n(X(\omega)) = f(X(\omega))$. Then it holds that $Y = f \circ X$.

$\square$

Remark. We don't know whether $X([Y = y_i]) \in \mathcal{R}^k$ or not. Otherwise, we can set $\bar H_i = X([Y = y_i])$ so that $X^{-1}(\bar H_i) = [Y = y_i]$ and the $\bar H_i, i \in [n]$ are disjoint automatically.

Independence

Definition. A finite or countably infinite sequence $X_1, X_2, \ldots$ of random variables are independent, if $\sigma(X_1), \sigma(X_2), \ldots$ are independent.

Theorem. The random variables $X_1, \dots, X_n$ are independent if and only if for all $x_i \in \mathbb{R}, i \in [n]$,

$$\Pr[ X_1 \le x_1, \ldots, X_n \le x_n ] = \Pr[ X_1 \le x_1 ] \cdots \Pr[ X_n \le x_n ].$$

Proof. Since the $\sigma$-field $\sigma(X_i)$ consists exactly of the sets $[X_i \in H]$ for $H \in \mathcal{R}$, $X_1, \dots, X_n$ are independent if and only if for all $H_i \in \mathcal{R}, i \in [n]$,

Since the intervals $(-\infty, x_i]$ forms a $\pi$-system generating $\mathcal{R}$, $[X_i \le x_i]$ forms a $\pi$-system generating $\sigma(X_i)$, which proves the theorem.

$\square$

Corollary. If the $X_i$ are simple random variables, then they are independent if and only if for all $x_i \in \mathbb{R}, i \in [n]$,

$$\Pr[ X_1 = x_1, \ldots, X_n = x_n ] = \Pr[ X_1 = x_1 ] \cdots \Pr[ X_n = x_n ].$$

Let $X = (X_1, \ldots, X_k)$ have distribution $\mu$ and distribution function $F$. Let $f_i : \mathbb{R}^k \rightarrow \mathbb{R}$ be the projection defined by $f_i(x_1,\ldots, x_k) = x_i$, $i \in [n]$. It is measurable, and therefore $X_i = f_i \circ X$ is measurable. Let the $X_i$ have distributions $\mu_i$ and distribution functions $F_i$ (the marginals). Consider the product measurable $\mu_1 \times \dots \times \mu_k$.

Theorem. The following conditions are equivalent

1. $X_i$ are independent if and only if
2. $$\mu = \mu_1 \times \dots \times \mu_k.$$
3. $$F(x_1, \ldots, x_k) = F(x_1) \dotsb F(x_k).$$

Proof.

Therefore, conditions (1) and (2) are equivalent.

Since the intervals $(-\infty, x_i]$ forms a $\pi$-system generating $\mathcal{R}$, $[X_i \le x_i]$ forms a $\pi$-system generating $\sigma(X_i)$, conditions (2) and (3) are equivalent.

$\square$

Theorem. Suppose that $\mu_i$ has density $f_i$. If the $X_i$ are independent, then $\mu$ has density

$$f(x) = f_1(x_1) \dotsb f_k(x_k).$$

Proof. Since the density function are non-negative, via Fubini's theorem, for each $H_1, \ldots, H_k \in \mathcal{R}$,

It follows by $\pi-\lambda$ theorem, for each $H \in \mathcal{R}^k$, we have

$\square$

Theorem. Let

$$\begin{matrix} X_{1, 1}, &X_{1, 2}, &\ldots \\ X_{2, 1}, &X_{2, 2}, &\ldots \\ \ldots &\ldots &\ldots \end{matrix}$$

be a (possibly countably infinite) sequence of independent random variables. The $\sigma$-fields $\mathcal{F}_n \doteq \sigma(X_{n, 1}, X_{n, 2}, \ldots )$, $n \in \mathbb{N}^+$ are independent.

Proof. For each $n \in \mathbb{N}^+$, define the class

which is the finite intersections of the sets $X_{n, k}^{-1} (H), k \in \mathbb{N}^+, H \in \mathcal{R}$. It is easy to see that $\emptyset \in \mathcal{A}_n$, and $\mathcal{A}_n$ is a $\pi$-system. Further,

• $\mathcal{A}_n$ generates $\mathcal{F}_n$, i.e., $\sigma(\mathcal{A}_n) = \mathcal{F}_n$:

  • $\sigma(\mathcal{A}_n) \subseteq \mathcal{F}_n$: via definition of $\mathcal{F}_n$, $X_{n, k}^{-1} (H) \in \mathcal{F}_n$, for each $k \in \mathbb{N}^+, H \in \mathcal{R}$. Hence, each element in $\mathcal{A}_n$ belongs to $\mathcal{F}_n$, and $\sigma(\mathcal{A}_n) \subseteq \mathcal{F}_n$.

  • $\sigma(\mathcal{A}_n) \supseteq \mathcal{F}_n$: via definition of $\mathcal{A}_n$, $X_{n, k}^{-1} (H) \in \mathcal{A}_n$, for each $k \in \mathbb{N}^+, H \in \mathcal{R}$. Hence, each $X_{n, k}$ is measurable $\sigma(\mathcal{A}_n)$. Via definition of $\mathcal{F}_n$, $\sigma(\mathcal{A}_n) \supseteq \mathcal{F}_n$.

• The $\mathcal{A}_n$ are independent: let $n_1, \ldots, n_m$ be a finite sequence of indexes in $\mathbb{N}^+$. Suppose that for each $j \in [m]$,

  $$B_{n_j} \doteq X_{n_j, i_1(j)}^{-1} (H_{i_1(j)}) \cap \ldots \cap X_{n_j, i_k(j)}^{-1} (H_{i_{k}(j)}).$$

  Then

  $$\begin{aligned} \Pr[ B_{n_1} \cap \ldots \cap B_{n_m} ] &= \Pr\left[ \cap_{j \in [m]} \left( X_{n_j, i_1(j)}^{-1} (H_{i_1(j)}) \cap \ldots \cap X_{n_j, i_k(j)}^{-1} (H_{i_{k}(j)}) \right) \right] \\ &= \prod_{j \in [m]} \Pr[ X_{n_j, i_1(j)} \in H_{i_1(j)}] \cdots \Pr[X_{n_j, i_k(j)} \in H_{i_{k}(j)} ] \\ &= \prod_{j \in [m]} \Pr[B_{n_j}]. \end{aligned}$$