Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(\Omega, \cal{A}, \mathbb{P})$ be a probability space and $X$ a random variable on $\Omega$. Let, also, $f:\Omega\to\mathbb{R}$ be a Borel function. Then:
$X$ and $f(X)$ are independent $\Longleftrightarrow$ there exists some $t\in\mathbb{R}$ such that $\mathbb{P}[f(X)=t]=1$, that is $f(X)$ is a degenerate r.v.

The only thing that I could make out is that if $X$ and $f(X)$ are independent, then
$\mathbb{P}[f(X)\in B]=0$ or $1$ for every Borel subset of $\mathbb{R}$, since $\sigma(f(X))\subseteq \sigma(X)$ and hence, $f(X)$ is independent of its self. Suppose, now, that $\mathbb{P}[f(X)\leq x]=0$ for all $x\in\mathbb{R}$. Then:
$\mathbb{P}[f(X)\in\mathbb{R}]=\mathbb{P}[\bigcup_{n=0}^{\infty}[f(X)\leq n]]\leq\sum_{n=0}^{\infty}[f(X)\leq n]=0$ which obviously is a contradiction since $\mathbb{P}[f(x)\in\mathbb{R}]=1$.

However, I don't know ow to prove this and my attempt isn't likely to become a complete solution.
Any help would be appreciated.
Thanks in advance!

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

If the random variable $Y$ is independent of itself, then there exists $y$ such that $Y=y$ almost surely.

To prove this, consider the CDF $F_Y:x\mapsto\mathbb P(Y\leqslant x)$ and note that for every $x$ the event $[Y\leqslant x]$ is independent of itself hence $F_Y(x)$ is $0$ or $1$. Since $F_Y$ has limits $0$ at $-\infty$ and $1$ at $+\infty$, the real number $y=\inf\{x\mid F_Y(x)=1\}$ is well defined and finite.

Since $F_Y$ is nondecreasing, $F(x)=0$ for every $x\lt y$ and $F_Y(x)=1$ for every $x\gt y$, hence $\mathbb P(y-u\lt Y\leqslant y+u)=1$ for every $u\gt0$, in particular, $\mathbb P(Y=y)=1$.

share|improve this answer
add comment

There is another proof of this fact, which follows immediately from the fact that you proved: $\Bbb P[f(x)\in B]\in \{0,1\}$ and reminds the Nested Intervals Theorem. For the shorthand let $$ \mu(B):=\Bbb P[f(X)\in B] $$ denote the distribution of $f(X)$.

So we have $\mu \in \{0,1\}$ and $\mu(\Bbb R) = 1$. Let us prove that it implies that $\mu$ is a degenerate distribution, i.e. there exists $x\in \Bbb R$ such that $\mu(\{x\}) = 1$.

  1. There exists a bounded closed interval $I_n =[-n,n]$ such that $\mu(I_n) = 1$. Indeed, if there is no such interval then $$ 1 = \mu(\Bbb R) = \mu\left(\bigcup_n I_n\right) = \lim_n \mu(I_n) = 0 $$ which is a contradiction.

  2. Now we construct a sequence of nested intervals. Denote $J_0 = I_n$, and let $$ J_0^{-} = [-n,0],\quad J_0^+ = [0,n]. $$ There is at least one of these intervals of measure $1$. Denote it by $J_1$.

  3. Repeat by induction the procedure for $J_k$ where $k = 1,2,\dots$: divide it symmetrically into two parts and put $J_{k+1}$ be any of these parts such that $\mu(J_{k+1}) = 1$.

  4. In the end, you have a decreasing sequence of compact intervals $J_0,\dots,J_k,\dots$ such that $\mathrm{diam}(J_k)\leq 2^{1-k}n $ thus $\bigcap_k J_k$ is some single point. We have $$ \mu\left(\bigcap_k J_k\right) = \lim_k \mu(J_k) = 1 $$ so that we found $x = \bigcap_k J_k$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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