5

Problem. Let $ X $ be a random variable, and $ f: \mathbb{R} \to \mathbb{R} $ a Borel function such that $ X $ and $ f(X) $ are independent. Show that there exists a constant $ c \in \mathbb{R} $ such that $ \mathsf{P}(f(X) = c) = 1 $.

This is what I have tried so far:

$ X $ and $ f(X) $ are independent, so $ X $ and $ X $ must be independent. I was then able to show that there exists a constant $ c \in \mathbb{R} $ such that $ \mathsf{P}(X = c) = 1 $. I’m not sure how to proceed further.

Berrick Caleb Fillmore
  • 5,269
Sahiba Arora
  • 10,847

1 Answers1

4

Here is an elementary solution.

Claim 1. For any Borel subsets $ A $ and $ B $ of $ \mathbb{R} $, we have $$ \mathsf{P}(f(X) \in A) \cdot \mathsf{P}(f(X) \in B) = \mathsf{P}(f(X) \in A \cap B). $$

Proof. Let $ A $ and $ B $ be Borel subsets of $ \mathbb{R} $. As $ X $ and $ f(X) $ are independent, we have \begin{align} \mathsf{P}(f(X) \in A) \cdot \mathsf{P}(f(X) \in B) & = \mathsf{P}(X \in {f^{\leftarrow}}[A]) \cdot \mathsf{P}(f(X) \in B) \\ & = \mathsf{P}(X \in {f^{\leftarrow}}[A] ~ \text{and} ~ f(X) \in B) \\ & = \mathsf{P}(f(X) \in A ~ \text{and} ~ f(X) \in B) \\ & = \mathsf{P}(f(X) \in A \cap B). \quad \blacksquare \end{align}

Claim 2. There exists an $ n \in \mathbb{Z} $ such that $ \mathsf{P}(f(X) \in [n,n + 1]) = 1 $.

Proof. Suppose otherwise. For each $ n \in \mathbb{Z} $, Claim 1 says that \begin{align} \mathsf{P}(f(X) \in [n,n + 1]) \cdot \mathsf{P}(f(X) \in \mathbb{R} \setminus [n,n + 1]) & = \mathsf{P}(f(X) \in [n,n + 1] \cap (\mathbb{R} \setminus [n,n + 1])) \\ & = \mathsf{P}(f(X) \in \varnothing) \\ & = 0. \end{align} Hence, for each $ n \in \mathbb{Z} $, we have $$ \mathsf{P}(f(X) \in [n,n + 1]) = 0 \qquad \text{or} \qquad \mathsf{P}(f(X) \in \mathbb{R} \setminus [n,n + 1]) = 0, $$ but as these two probabilities add up to $ 1 $, we find that $ \mathsf{P}(f(X) \in [n,n + 1]) = 0 $. Consequently, $$ 1 = \mathsf{P}(f(X) \in \mathbb{R}) = \mathsf{P} \! \left( f(X) \in \bigcup_{n \in \mathbb{Z}} [n,n + 1] \right) \leq \sum_{n \in \mathbb{Z}} \mathsf{P}(f(X) \in [n,n + 1]) = \sum_{n \in \mathbb{Z}} 0 = 0, $$ which is a contradiction. $ \quad \blacksquare $

Claim 3. Let $ a,b \in \mathbb{R} $ with $ a < b $. If $ \mathsf{P}(f(X) \in [a,b]) = 1 $, then $$ \mathsf{P} \! \left( f(X) \in \left[ a,\frac{a + b}{2} \right] \right) = 1 \qquad \text{or} \qquad \mathsf{P} \! \left( f(X) \in \left[ \frac{a + b}{2},b \right] \right) = 1. $$

Proof. By way of contradiction, suppose that $ \mathsf{P}(f(X) \in [a,b]) = 1 $ but $$ \mathsf{P} \! \left( f(X) \in \left[ a,\frac{a + b}{2} \right] \right) \neq 1 \qquad \text{and} \qquad \mathsf{P} \! \left( f(X) \in \left[ \frac{a + b}{2},b \right] \right) \neq 1. $$ By Claim 1, we have \begin{align} \mathsf{P} \! \left( f(X) \in \left[ a,\frac{a + b}{2} \right] \right) \cdot \mathsf{P} \! \left( f(X) \in \left( \frac{a + b}{2},b \right] \right) & = \mathsf{P} \! \left( f(X) \in \left[ a,\frac{a + b}{2} \right] \cap \left( \frac{a + b}{2},b \right] \right) \\ & = \mathsf{P}(f(X) \in \varnothing) \\ & = 0. \end{align} Hence, $$ \mathsf{P} \! \left( f(X) \in \left[ a,\frac{a + b}{2} \right] \right) = 0 \qquad \text{or} \qquad \mathsf{P} \! \left( f(X) \in \left( \frac{a + b}{2},b \right] \right) = 0. $$ However, these two probabilities add up to $ 1 $, so $ \mathsf{P} \! \left( f(X) \in \left[ a,\dfrac{a + b}{2} \right] \right) = 0 $. A similar argument yields $ \mathsf{P} \! \left( f(X) \in \left[ \dfrac{a + b}{2},b \right] \right) = 0 $ as well. Therefore, \begin{align} 1 & = \mathsf{P}(f(X) \in [a,b]) \\ & = \mathsf{P} \! \left( f(X) \in \left[ a,\frac{a + b}{2} \right] \cup \left[ \frac{a + b}{2},b \right] \right) \\ & \leq \mathsf{P} \! \left( f(X) \in \left[ a,\frac{a + b}{2} \right] \right) + \mathsf{P} \! \left( f(X) \in \left[ \frac{a + b}{2},b \right] \right) \\ & = 0 + 0 \\ & = 0, \end{align} which is a contradiction. $ \quad \blacksquare $

Using Claims 2 and 3, we can define a nested sequence $ (I_{n})_{n \in \mathbb{N}} $ of closed intervals such that

  • $ \mathsf{P}(f(X) \in I_{n}) = 1 $ for every $ n \in \mathbb{N} $, and
  • $ \displaystyle \lim_{n \to \infty} |I_{n}| = 0 $.

By the Nested-Intervals Theorem, there exists a real number $ c $ such that $ \displaystyle \bigcap_{n = 1}^{\infty} I_{n} = \{ c \} $, so $$ \mathsf{P}(f(X) = c) = \mathsf{P} \! \left( f(X) \in \bigcap_{n = 1}^{\infty} I_{n} \right) = \lim_{n \to \infty} \mathsf{P}(f(X) \in I_{n}) = \lim_{n \to \infty} 1 = 1. $$

Berrick Caleb Fillmore
  • 5,269