15
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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 $y\ge k$.

$\endgroup$
10
$\begingroup$

$X$ 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 $y\notin B$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.