When Is a Random Variable Almost Surely Constant?

A real-valued random variable is almost surely constant if there is a fixed number $c$ such that $P(X=c)=1$. Its value may differ from $c$ on a probability-zero set; that does not affect the claim.

A standard result is that if $X$ is independent of itself, then it is almost surely constant. The proof uses events such as $\{X\leq q\}$ for rational numbers $q$. The rational thresholds are enough to pin down the distribution without relying on an uncountable collection of events.

What “independent of itself” means

Two random variables are independent when events specified by their values are independent. Thus, for every pair of Borel sets $B,C\subseteq\mathbb{R}$,

$$P(X\in B, X\in C)=P(X\in B)P(X\in C).$$

When both variables are the same $X$, take $B=C$. The event on the left is just $\{X\in B\}$, so

$$P(X\in B)=P(X\in B)^2.$$

The only probabilities $p$ satisfying $p=p^2$ are $0$ and $1$. Therefore every event determined by $X$ has probability either zero or one. In the language of sigma-algebras, $\sigma(X)$ is independent of itself.

Proof using rational thresholds

Define the distribution function $F(t)=P(X\leq t)$. For every rational number $q$, the event $\{X\leq q\}$ is determined by $X$. The preceding argument gives

$$F(q)\in\{0,1\}\qquad\text{for every }q\in\mathbb{Q}.$$

Because $X$ is real-valued, it is finite with probability one. Consequently, $F(q)$ is zero for sufficiently small rational thresholds and one for sufficiently large ones. Define

$$c=\inf\{q\in\mathbb{Q}:F(q)=1\}.$$

This is a finite real number. If $q<c$, then $F(q)=0$ by the definition of the infimum. If $q>c$, there is a rational $r$ with $c<r<q$ and $F(r)=1$; since $F$ is nondecreasing, $F(q)=1$ as well.

These facts imply that $X=c$ almost surely. For example, choose rational numbers $q_n<c$ increasing to $c$. Each event $\{X\leq q_n\}$ has probability zero, and their union is $\{X<c\}$, so $P(X<c)=0$. Choose rational numbers $r_n>c$ decreasing to $c$. Each event $\{X\leq r_n\}$ has probability one, and their intersection is $\{X\leq c\}$, so $P(X\leq c)=1$. Together, these give $P(X=c)=1$.

The converse also holds: if $X=c$ almost surely, then events determined by $X$ have probability zero or one, and $X$ is independent of itself. Null events in $\sigma(X)$ do not change that conclusion.

Worked example: a Bernoulli random variable

Suppose $X$ is Bernoulli with $P(X=1)=p$ and $P(X=0)=1-p$. If $X$ is independent of itself, the event $\{X=1\}$ must be independent of itself. Hence

$$P(X=1)=P(X=1)^2, \qquad p=p^2.$$

So $p=0$ or $p=1$. In either case, $X$ is constant with probability one: it equals $0$ when $p=0$ and equals $1$ when $p=1$. Any value $0<p<1$ rules out independence of $X$ from itself.

Other common ways to establish constancy

Self-independence is only one criterion. If $X$ has a finite second moment, then

$$\operatorname{Var}(X)=\mathbb{E}\big[(X-\mathbb{E}[X])^2\big].$$

If this variance is zero, the nonnegative random variable $(X-\mathbb{E}[X])^2$ has expectation zero, so it is zero almost surely. Thus $X=\mathbb{E}[X]$ almost surely. Conversely, an almost surely constant variable with finite second moment has variance zero.

More generally, if $g$ is a nonnegative measurable function and $\mathbb{E}[g(X)]=0$, then $g(X)=0$ almost surely. For example, taking $g(x)=(x-c)^2$ proves that a finite second moment around $c$ equal to zero forces $X=c$ almost surely.

Common mistakes

• Confusing self-independence with an independent copy. An independent copy $Y$ of $X$ can have the same non-degenerate distribution as $X$. The result concerns $X$ and $X$ itself, not $X$ and a separately constructed copy.
• Thinking equal distributions imply independence. Two random variables may have the same distribution and still be dependent. A variable and itself are generally dependent unless it is almost surely constant.
• Ignoring “almost surely.” The result does not require $X=c$ at every point of the sample space; exceptional outcomes with probability zero are allowed.
• Applying the zero-variance criterion without its assumptions. The usual variance argument requires a finite second moment. Self-independence, by contrast, gives the conclusion for any real-valued random variable, whether or not its expectation or variance exists.