Probability Theory Solution – Problem 58: Prove that a random variable is independent of itself if and only if is almost surely…

Question

Prove that a random variable $X$ is independent of itself if and only if $X$ is almost surely constant.

Step-by-step solution

Proof: Step 1. If $X$ is almost surely constant: Suppose $P(X = c) = 1$ . We need to show that $X$ is independent of itself, i.e., for any Borel sets $A, B \subseteq \mathbb{R}$ , $P(X \in A, X \in B) = P(X \in A) \cdot P(X \in B).$ Case 1: $c \in A \cap B$ Then $P(X \in A) = 1,\ P(X \in B) = 1,\ P(X \in A,\ X \in B) = 1$ , so $1 = 1 \times 1$ holds. Case 2: $c \in A$ but $c \notin B$ Then $P(X \in A) = 1,\ P(X \in B) = 0,\ P(X \in A,\ X \in B) = 0$ , which holds. Case 3: $c \notin A$ but $c \in B$ , similar. Case 4: $c \notin A$ and $c \notin B$ Then $P(X \in A) = 0,\ P(X \in B) = 0,\ P(X \in A,\ X \in B) = 0$ , which holds. Therefore, "almost surely constant" implies $X$ is independent of itself.

Step 2. If $X$ is independent of itself: Independence means that for any Borel set $A, B$ : $P(X \in A, X \in B) = P(X \in A) P(X \in B).$ In particular, taking $A = B$ , we have $P(X \in A) = \big[P(X \in A)\big]^2.$ Thus for any Borel set $A$ , $P(X \in A) \in \{0, 1\}.$

Step 3. We show that $\{P(X \in A) \in \{0, 1\}\ \forall A \in \mathcal{B}(\mathbb{R})\}$ implies $X$ is almost surely constant. Define the distribution function $F(t) = P(X \le t)$ . By the previous step, $F(t) \in \{0, 1\}$ . Moreover, $F$ is monotone non-decreasing and right-continuous, with $\lim_{t \to -\infty} F(t) = 0$ and $\lim_{t \to +\infty} F(t) = 1$ . Therefore there exists a unique $c \in \mathbb{R}$ such that: For $t < c$ , $F(t) = 0$ , so $P(X \le t) = 0$ , hence $P(X > t) = 1$ . For $t \ge c$ , $F(t) = 1$ . Taking $t = c - 1/n$ , we get $P(X > c - 1/n) = 1$ , so $P(X \ge c) = 1$ (by taking the intersection). Also, taking $t = c$ , we get $P(X \le c) = 1$ . Therefore $P(X = c) = P(X \le c) - P(X < c)$ . Note that $P(X < c) = \lim_{t \uparrow c} F(t) = 0$ , so $P(X = c) = 1 - 0 = 1.$

Step 4. In summary, $X$ is independent of itself implies the distribution function takes only values 0 or 1, which implies there exists $c$ with $P(X=c)=1$ ; conversely, if $P(X=c)=1$ then clearly $X$ is independent of itself. QED.

Final answer

QED.

Marking scheme

The following is the marking scheme based on the official solution (maximum 7 points):

1. Checkpoints (max 7 pts)

Part 1: Sufficiency proof (2 points)

*Prove that " $X$ almost surely constant $\Rightarrow$ $X$ is independent of itself"*

[1 pt] State that if $P(X=c)=1$ , then for any Borel set $A$ , the probability $P(X \in A)$ takes only the values $0$ or $1$ (depending on whether $c \in A$ ).
[1 pt] Verify the independence definition: show that $P(X \in A, X \in B) = P(X \in A)P(X \in B)$ holds for all $0/1$ combinations (i.e., $1\times 1=1, 1\times 0=0$ , etc.).

Part 2: Necessity proof (5 points)

*Prove that " $X$ independent of itself $\Rightarrow$ $X$ almost surely constant"*

Score exactly one chain among the following; if multiple chains appear, take the highest score without adding across chains.

> Chain A: Via the 0-1 property and the distribution function (standard approach)

> - [2 pts] Derive the 0-1 property: Using the independence definition (setting $B=A$ or similar), derive $P(X \in A) = P(X \in A)^2$ , and conclude $P(X \in A) \in \{0, 1\}$ .

> - [2 pts] Locate the constant $c$ : Using the monotonicity, right-continuity, and limit properties of the distribution function $F(t)$ (which takes only values $0$ and $1$ ), show there exists a unique jump point $c$ (or use the supremum principle $c = \sup\{t: F(t)=0\}$ ).

> - [1 pt] Confirm probability concentrates at $c$ : Rigorously show the jump has magnitude 1, i.e., $P(X=c) = F(c) - F(c-) = 1 - 0 = 1$ (or equivalent set-theoretic argument).

> Chain B: Via variance/moment properties (alternative approach)

> - [1 pt] Construct bounded variable / justify moment existence: Introduce a bounded function (e.g., $Y=\arctan X$ ) or truncation, or explicitly discuss moment existence here.

> - [2 pts] Derive zero variance: Use independence to derive $\text{Var}(Y) = \text{Cov}(Y, Y) = 0$ or $E[Y^2] = (E[Y])^2$ .

> - [1 pt] Conclude almost surely constant: From zero variance, deduce $Y$ is almost surely constant.

> - [1 pt] Transfer the conclusion: By properties of the transformation (e.g., injectivity), show the original variable $X$ is almost surely constant.

Total (max 7)

2. Zero-credit items

Merely copying the definitions of "independent" or "almost surely constant" without any targeted derivation.
Simply asserting "a constant variable is obviously independent" or "an independent variable is obviously constant" without mathematical justification.
Confusing the concept of $X$ independent of itself with $X$ independent of another variable $Y$ , rendering the derivation invalid.

3. Deductions

*Apply the most severe single deduction below (minimum total score is 0):*

[Cap at 3/7] Restrictive distributional assumption: In the necessity proof, assuming without justification that $X$ is discrete (enumerating probability masses) or continuous (assuming a density $f(x)$ exists), thereby losing generality.
[Cap at 5/7] Failure to justify moment existence: If using Chain B, directly computing $Var(X)=0$ without verifying that $X$ has a finite second moment (or without using a bounded transformation), the argument is considered insufficiently rigorous.
[-1 pt] Logical gap: In Chain A, after obtaining $F(t) \in \{0,1\}$ , directly asserting "therefore $F(t)$ is a step function" without using distribution function properties (such as monotonicity or limits) as intermediate justification.

Probability Theory – Problem 58: Prove that a random variable is independent of itself if and only if is almost surely…