# Independence between a constant random variable and another random variable.

Intuitively, I understand that if $Y$ is a constant random variable and $X$ is another random variable, then $X$ and $Y$ are independent.

However, I can't make a formal proof because I can't show that their joint density function are the product of two functions that rely only on x's and y's respectively or using similar methods.

(What is the density function of a constant random variable for example?)

Can you give me a hint in order to make a proof?

Hint: Work with the cumulative distribution functions. Show that for all $x$ and $y$ we have $\Pr(X\le x\cap Y\le y)=\Pr(X\le x)\Pr(Y\le y)$.

Note that if $Y=k$ with probability $1$, then $F_Y(y)=0$ if $y\lt k$, and $F_Y(y)=1$ if $y\ge k$.

$X$ and $Y$ are independent if and only if $P(X\in A, Y\in B)=P(X\in A)P(Y\in B)$ for all $A,B$.

Assume $Y=y$ for some $y\in\mathbb{R}$. Then $$P(X\in A, Y\in B)=\left\{\begin{array}{ll}0&\mathrm{if}\,\,y\notin B,\\ P(X\in A)&\mathrm{if}\,\,y\in B\end{array}\right.$$

But notice $P(Y\in B)=1$ if $y\in B$ and $P(Y\in B)=0$ if $y\notin B$.

Here is a fun proof using (introductory) measure theory. $$\newcommand{\ind}{\perp\kern-5pt\perp}$$

Short version

Let $$(\Omega, \mathcal{F}, P)$$ be a probability space, $$C : \Omega \to \Psi$$ a constant random variable, $$X: \Omega \to \Psi$$ an arbitrary random variable, and $$\sigma_X$$ the $$\sigma$$-field generated by $$X$$. Note that $$\sigma_C=\{\emptyset, \Omega \}$$. But since $$\Omega$$ and $$\emptyset$$ are independent of any other event in $$\mathcal{F}$$, we have $$\sigma_C \ind \sigma_X$$. Therefore, $$C \ind X$$.

Long version

(Assumes almost$$^\star$$ no prior knowledge of measure theory.) First, given some probability space $$(\Omega, \mathcal{F}, P)$$, note that $$\Omega$$ and $$\emptyset$$ are independent of any other event $$A \in \mathcal{F}$$. To see this, consider the definition of independent events, which is that $$A \ind B$$ if $$P(A \cap B) = P(A)P(B)$$. Now, observe that, $$P(A \cap \Omega) = P(A) = P(A) \cdot 1 = P(A) P(\Omega)$$, so $$\Omega$$ is independent of any event in $$\mathcal{F}$$. A similar argument holds for $$\emptyset$$.

Next, note that if $$C : \Omega \to \Psi$$ is a constant random variable, then $$\sigma_C$$, the $$\sigma$$-field generated by $$C$$, is trivial. In other words, $$\sigma_C = \{ \emptyset, \Omega \}$$. To see this, consider that the definition of a sigma field generated by random variable C is $$\sigma_C := \{ \{ C^{-1}(B) \}: B \in \mathcal{B}(\Psi) \}$$, where $$\mathcal{B}(\Psi)$$ are the Borel sets of $$\Psi$$. Then note that if $$C$$ takes on constant value $$c_0 \in \Psi$$, then $$C^{-1}(B) = \Omega$$ if $$c_o \in B$$, and otherwise $$C^{-1}(B) = \emptyset$$.

Now note that $$\sigma_C$$ and $$\sigma_X$$ must be independent $$\sigma$$-fields for any random variable $$X$$. To see this, consider that two $$\sigma$$-fields $$\mathcal{F}$$ and $$\mathcal{G}$$ are defined to be independent if events $$F$$ and $$G$$ are independent for any $$F \in \mathcal{F}, G \in \mathcal{G}$$. We want to show this is true when $$\mathcal{F} = \sigma_X, \mathcal{G}=\sigma_C$$. But we have already determined that $$\sigma_C = \{ \Omega, \emptyset \}$$, and that events $$\Omega, \emptyset$$ are independent from all other events (including events in $$\sigma_X$$). So we are done.

Finally, note that two random variables X,Y are independent if the $$\sigma$$-fields generated by them are independent. In other words, $$\sigma_X \ind \sigma_Y \implies X \ind Y$$. To see this, recall the definition of $$X \ind Y$$, which is $$\forall B, B' \in \mathcal{B}(\Psi), \{ X^{-1}(B)\} \ind \{Y^{-1}(B')\}$$. But by construction, $$\{ X^{-1}(B)\} \in \sigma_X$$ and $$\{Y^{-1}(B')\} \in \sigma_Y$$, and those events are independent by assumption.

Footnotes

$$\star$$: The only prerequisites are (1) the definition of a sigma field and (2) Borel sets. The former is introductory and can be looked up. For some sense of the latter, simply consider $$\mathcal{B}(\mathbb{R})$$, which is the smallest sigma field that contains all the intervals.

Show a more general result, that if $$Y$$ is a constant random variable with value $$c$$ with probability 1, it is independent to any random variable $$X$$. Essentially, the result we need is from random statement 24. Let $$k \neq c$$ be a constant.

$$\Pr(Y=c)=1$$, so $$\Pr (B, Y=c) = \Pr (B)$$ for all events $$B$$. In particular, setting $$B=\{ X=x \}$$ gives $$\Pr (X=x) \Pr (Y=c) = \Pr (X=x) = \Pr(X = x, Y = c)$$.

$$\Pr(Y=k)=0$$, so $$\Pr (B \cup (Y=k)) = \Pr (B)$$ for all events $$B$$. In particular, setting $$B=\{ X=x \}$$ and using the inclusion-exclusion rule gives $$\Pr(X = x, Y = k) = \Pr (X=x) + \Pr (Y=k) - \Pr (X=x \cup Y=k) = \Pr (X=x) + 0 - \Pr (X=x) = 0 = \Pr (X=x) \Pr(Y=k)$$.