Independence of random variables (incomplete)

Dependencies: (incomplete)

$\newcommand{\Fcal}{\mathcal{F}}$ $\newcommand{\Ecal}{\mathcal{E}}$ $\newcommand{\Tcal}{\mathcal{T}}$ Let $(\Omega, \Fcal, \Pr)$ be a probability space. Let $(D, \Ecal)$ and $(D', \Ecal')$ be $\sigma$-algebras. Let $X$ and $Y$ be random variables from $(\Omega, \Fcal, \Pr)$ to $(D, \Ecal)$ and $(D', \Ecal')$ respectively.

Let $\Tcal \subseteq \Ecal$ and $\Tcal' \subseteq \Ecal'$.

$X$ is said to be $\Tcal$-independent of $A \in \Fcal$ iff $\forall B \in \Tcal$, the events $X \in B$ and $A$ are independent. $X$ is said to be independent of $A$ iff $X$ is $\Ecal$-independent of $A$.

$X$ is said to be $(\Tcal, \Tcal')$-independent of $Y$ iff $\forall B \in \Tcal$ and $\forall B' \in \Tcal'$, the events $X \in B$ and $Y \in B'$ are independent. $X$ is said to be independent of $Y$ iff $X$ is $(\Ecal, \Ecal')$-independent of $Y$.

We say that $A \in \Fcal$ is $\Tcal$-independent of $X$ iff $X$ is $\Tcal$-independent of $A$. It is easy to prove that $X$ is $(\Tcal, \Tcal')$-independent of $Y$ iff $Y$ is $(\Tcal', \Tcal)$-independent of $X$, since independence of events is symmetric.

Equivalent definition for discrete random variable

Suppose $D$ is countable and $\Ecal$ is the power-set of $D$. So $X$ is discrete. Let $\Tcal = \{\{x\}: x \in D_1\}$. Let $\Tcal' \subseteq \Ecal'$ be an arbitrary set.

Theorem: $X$ is independent of $A \in \Fcal$ iff $X$ is $\Tcal$-independent of $A$. $X$ is $(\Ecal, \Tcal')$-independent of $Y$ iff $X$ is $(\Tcal, \Tcal')$-independent of $Y$.

Proof: Let $S \in \Ecal$ and $S' \in \Tcal'$. $X \in S = \bigcup_{x \in S} (X = x)$. $\forall x \in D$, the event $X = x$ is independent of $A$ and $Y \in S'$ because $X$ is independent of $A$ and $Y$. Since independence is preserved by countable disjoint unions, $X \in S$ is independent of $A$ and $Y \in S'$.

As a corollary, if $X$ and $Y$ are both discrete, then they are independent iff $f_{X,Y}(x, y) = f_X(x)f_Y(y)$, i.e. their joint probability mass function is equal to the product of their mass functions.

Equivalent definitions for real-valued random variables

Suppose $D = \mathbb{R}$ and $\Tcal = \{(-\infty, x]: x \in \mathbb{R}\}$ $\Ecal = \sigma(\Tcal) = \mathcal{B}(\mathbb{R})$. Let $\Tcal' \subseteq \Ecal'$ be an arbitrary set.

Theorem: $X$ is independent of $A \in \Fcal$ iff $X$ is $\Tcal$-independent of $A$. $X$ is $(\Ecal, \Tcal')$-independent of $A$ iff $X$ is $(\Tcal, \Tcal')$-independent of $Y$.

Proof: (Pending)

As a corollary, if $X$ and $Y$ are real-valued, then they are independent iff $F_{X,Y}(x, y) = F_X(x)F_Y(y)$, i.e. their joint CDF is equal to the product of their CDFs.

Joint density function

We can prove that when $X$ and $Y$ are continuous, they are independent iff their joint density function is the product of their density functions, i.e. $f_{X, Y}(x, y) = f_X(x)f_Y(y)$.

\[ \int_{-\infty}^x \int_{-\infty}^y f_X(x)f_Y(y) dy dx = \left(\int_{-\infty}^x f_X(x) dx \right)\left(\int_{-\infty}^y f_Y(y) dy\right) = F_X(x)F_Y(y) \tag{definition of density} \]

Therefore, if $f_{X, Y}(x, y) = f_X(x)f(y)$, then \[ F_{X, Y}(x, y) = \int_{-\infty}^x \int_{-\infty}^y f_{X, Y}(x, y) = \int_{-\infty}^x \int_{-\infty}^y f_X(x)f_Y(y) = F_X(x)F_Y(y) \] so $X$ and $Y$ are independent.

If $X$ and $Y$ are independent, \[ \int_{-\infty}^x \int_{-\infty}^y f_X(x)f_Y(y) = F_X(x)F_Y(y) = F_{X, Y}(x, y) \] So $f_X(x)f_Y(y)$ is a joint density function for $(X, Y)$.

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Depth: 7
Number of transitive dependencies: 15

Transitive dependencies:

/sets-and-relations/de-morgan-laws
/sets-and-relations/countable-set
/analysis/topological-space
σ-algebra
σ-algebra is closed under countable intersections
Measure
Probability
Conditional probability (incomplete)
Independence of events
Independence of composite events
Generated σ-algebra
Measurable function
Borel algebra
Generators of the real Borel algebra (incomplete)
Random variable