Probability Theory Solution – Problem 28: 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. Suppose $X$ is almost surely constant. Let $P(X = c) = 1$ . We wish to show that $X$ is independent of itself, i.e., 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 $c \in B$ . Analogous to Case 2. 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. Suppose $X$ 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, taking $A = B$ , 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. 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 nondecreasing 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 $P(X > c - 1/n) = 1$ , and by countable intersection $P(X \ge c) = 1$ . At the same time, taking $t = c$ 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 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 grading 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 $P(X \in A)$ takes only the values $0$ 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."*

Grade whichever of the following logical paths the student uses (if multiple paths appear, take the highest score; do not combine scores across paths):

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

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

> - [2 pts] Locate the constant value $c$ : Using the monotonicity, right-continuity, and limiting behavior of the distribution function $F(t)$ (which takes only 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 argue the jump has magnitude 1, i.e., $P(X=c) = F(c) - F(c-) = 1 - 0 = 1$ (or an equivalent set-theoretic argument).

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

> - [1 pt] Construct a bounded variable / justify moment existence: Introduce a bounded function (e.g., $Y=\arctan X$ ) or a truncation, or explicitly address moment existence at this point. *(If the student directly assumes $E[X^2] < \infty$ without justifying generality, see the deductions below.)*

> - [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: From zero variance, deduce that $Y$ is almost surely constant.

> - [1 pt] Transfer the conclusion: Use properties of the transformation (e.g., injectivity) to 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" from the problem statement without any substantive derivation.
Simply asserting "a constant random variable is obviously independent" or "an independent-of-itself variable is obviously constant" without providing 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 item below (minimum 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 probability density function $f(x)$ exists), thereby losing generality.
[Cap at 5/7] Failure to justify moment existence: If using Path B, directly computing $\text{Var}(X)=0$ without verifying that $X$ has a finite second moment (or without using a bounded transformation); this constitutes 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 (such as monotonicity or limiting behavior).

Probability Theory – Problem 28: Prove that a random variable is independent of itself if and only if is almost surely…