Probability Theory Solution – Problem 37: 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 $P(X = c) = 1$ . To show that $X$ is independent of itself, we must verify that for all 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 $1 = 1 \times 1$ . The identity holds. Case 2: $c \in A$ but $c \notin B$ Then $P(X \in A) = 1,\ P(X \in B) = 0,\ P(X \in A,\ X \in B) = 0$ . The identity holds. Case 3: $c \notin A$ but $c \in B$ . The argument is analogous. Case 4: $c \notin A$ and $c \notin B$ Then $P(X \in A) = 0,\ P(X \in B) = 0,\ P(X \in A,\ X \in B) = 0$ . The identity holds. Therefore, "almost surely constant" implies that $X$ is independent of itself. Step 2. If $X$ is independent of itself Independence means that for all Borel sets $A, B$ : $P(X \in A, X \in B) = P(X \in A) P(X \in B).$ In particular, setting $A = B$ , we obtain $P(X \in A) = \big[P(X \in A)\big]^2.$ Hence for every Borel set $A$ , $P(X \in A) \in \{0, 1\}.$ Step 3. $\{P(X \in A) \in \{0, 1\}\ \forall A \in \mathcal{B}(\mathbb{R})\}$ implies that $X$ 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 monotonically nondecreasing and right-continuous, satisfying $\lim_{t \to -\infty} F(t) = 0$ and $\lim_{t \to +\infty} F(t) = 1$ . Therefore there exists a unique $c \in \mathbb{R}$ such that: For $t < c$ , $F(t) = 0$ , i.e., $P(X \le t) = 0$ , hence $P(X > t) = 1$ . For $t \ge c$ , $F(t) = 1$ . Setting $t = c - 1/n$ , we get $P(X > c - 1/n) = 1$ , and by taking the intersection, $P(X \ge c) = 1$ . Setting $t = c$ gives $P(X \le c) = 1$ . 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, $X$ independent of itself implies that the distribution function takes only the values 0 and 1, which in turn implies the existence of $c$ with $P(X=c)=1$ ; conversely, if $P(X=c)=1$ , then $X$ is clearly independent of itself. QED.

Final answer

QED.

Marking scheme

The following is an undergraduate mathematics marking rubric based on the official solution (total: 7 points):

1. Checkpoints (max 7 pts)

Part I: Sufficiency Proof [2 pts total]

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

[1 pt] State that if $P(X=c)=1$ , then for every Borel set $A$ , the probability $P(X \in A)$ equals either $0$ or $1$ (depending on whether $c \in A$ ).
[1 pt] Verify the independence identity: 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 II: Necessity Proof [5 pts total]

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

Award marks for any ONE of the following logical paths (if multiple paths appear, take the highest score; do not combine):

> Path A: Via the 0–1 property and the distribution function (official solution approach)

> - [2 pts] Derive the 0–1 property: Using the independence definition (set $B=A$ or an analogous argument), obtain $P(X \in A) = P(X \in A)^2$ , and conclude that $P(X \in A) \in \{0, 1\}$ .

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

> - [1 pt] Confirm that probability concentrates at $c$ : Rigorously argue that the jump size equals 1, i.e., $P(X=c) = F(c) - F(c-) = 1 - 0 = 1$ (or provide an equivalent set-theoretic argument).

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

> - [1 pt] Introduce a bounded transform or justify moment existence: Introduce a bounded function (e.g., $Y=\arctan X$ ) or a truncation, or explicitly address the existence of moments. *(If $E[X^2] < \infty$ is assumed without justification of generality, see deductions.)*

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

> - [1 pt] Conclude almost sure constancy: Deduce from zero variance that $Y$ is almost surely constant.

> - [1 pt] Transfer the conclusion: Use properties of the transform (e.g., injectivity) to conclude that the original variable $X$ is almost surely constant.

Total (max 7)

2. Zero-Credit Items

Merely restating the definitions of "independence" or "almost surely constant" from the problem statement without any substantive derivation.
Simply asserting "a constant random variable is obviously independent of itself" or "an independent-of-itself random variable is obviously constant" without mathematical justification.
Confusing independence of $X$ with itself and independence of $X$ and $Y$ , rendering the derivation invalid.

3. Deductions

*Apply the most severe applicable deduction (minimum score is 0):*

[Cap at 3/7] Unjustified restriction to a specific distribution type: In the necessity proof, assuming without justification that $X$ is discrete (enumerating probability masses) or continuous (assuming a probability density function $f(x)$ exists), thereby losing generality.
[Cap at 5/7] Failure to justify moment existence: If Path B is used, computing $\text{Var}(X)=0$ directly without verifying that $X$ has a finite second moment (or without using a bounded transform) is considered a gap in rigor.
[-1 pt] Logical gap: In Path A, after obtaining $F(t) \in \{0,1\}$ , directly asserting "therefore $F(t)$ is a step function" without invoking properties of the distribution function (e.g., monotonicity or limits).

Probability Theory – Problem 37: Prove that a random variable is independent of itself if and only if is almost surely…