Support and Tail Limits

Why Tail Limits Need an Endpoint Interpretation

For every real-valued random variable, the probability of exceeding a threshold tends to zero as the threshold moves far enough to the right. That fact alone does not tell us whether the variable has a finite upper bound: a distribution can assign positive probability beyond every finite threshold and still have a tail that tends to zero. To distinguish these situations, we connect tail probabilities to the support and its endpoints.

The distinction is useful when reading a distribution from its CDF or survival function. If the distribution has a finite upper endpoint, the survival function eventually becomes exactly zero. If its upper support is unbounded, the survival function remains positive at every finite threshold, even though its limit at infinity is zero.

Support and Its Endpoints

Definition: The support of the distribution of a real-valued random variable $X$, written $\operatorname{supp}(X)$, is the set of points $x\in\mathbb{R}$ such that every open interval containing $x$ has positive probability under the distribution. Its upper and lower endpoints are $$ b=\sup \operatorname{supp}(X), \qquad a=\inf \operatorname{supp}(X), $$ where either endpoint may be $+\infty$ or $-\infty$, respectively. A finite $b$ is a finite upper endpoint; a finite $a$ is a finite lower endpoint.

Support is a property of the distribution, not a statement that the random variable must equal an endpoint with positive probability. For example, a continuous distribution can have a finite upper endpoint while assigning probability zero to that endpoint. The support is closed, so a finite endpoint belongs to the support, even when the probability of the single endpoint is zero.

For the CDF $F(x)=P(X\le x)$, define the strict upper-tail probability, or survival function, by $$ \overline F(x)=P(X>x)=1-F(x). $$ The strict inequality matters at atoms: $P(X>x)$ does not include probability located exactly at $x$.

Recognizing a Finite Upper Endpoint

Theorem: The distribution of $X$ has a finite upper endpoint $b$ if and only if $F(c)=1$ for at least one finite real number $c$. In that case, $$ b=\inf\{x\in\mathbb{R}:F(x)=1\}, $$ and $F(x)<1$ for $x<b$, while $F(x)=1$ for $x\ge b$. Equivalently, $\overline F(x)>0$ for $x<b$ and $\overline F(x)=0$ for $x\ge b$.

This gives a direct diagnostic from the CDF: look for the first point from which the CDF stays at one. A finite upper endpoint is not merely a point where the CDF is close to one. It is the threshold beyond which the probability of exceeding the threshold is exactly zero.

Proof. First suppose the support has finite supremum $b$. The distribution assigns probability one to its support, so $X\le b$ with probability one. Thus $F(b)=1$, and the set $\{x:F(x)=1\}$ is nonempty. If $x<b$, the definition of supremum gives a support point $y>x$. There is an open interval around $y$ lying entirely to the right of $x$, and that interval has positive probability by the definition of support. Therefore $P(X>x)>0$, so $F(x)<1$. For every $x\ge b$, $X\le b\le x$ almost surely, giving $F(x)=1$. Hence the infimum of the set where $F$ equals one is $b$.

Conversely, suppose $F(c)=1$ for some finite $c$. Then $P(X>c)=0$. No point strictly greater than $c$ can be in the support: a sufficiently small open interval around such a point lies entirely above $c$ and has probability zero. Thus the support is bounded above, and its supremum $b$ is finite. Applying the first part identifies $b$ as the infimum of the points where $F$ equals one and gives the stated CDF and survival-function behavior. $\square$

Worked Example: A Continuous Distribution With a Finite Endpoint

Let $X$ have a uniform distribution on $[-2,3]$. Its CDF is $$ F(x)= \begin{cases} 0, & x<-2,\\ \dfrac{x+2}{5}, & -2\le x<3,\\ 1, & x\ge 3. \end{cases} $$

For every $x<3$ in the interval $[-2,3)$, $(x+2)/5<1$; for $x<-2$, the CDF is zero. The first point from which $F(x)=1$ is therefore $b=3$. The support is $[-2,3]$, with finite upper endpoint $3$.

The strict survival function is $$ \overline F(x)= \begin{cases} 1, & x<-2,\\ \dfrac{3-x}{5}, & -2\le x<3,\\ 0, & x\ge 3. \end{cases} $$ For instance, at $x=2$, $\overline F(2)=(3-2)/5=1/5$; at $x=3$, it is zero. Since this distribution has no atom at $3$, the survival probability also tends to zero as $x$ approaches $3$ from below.

Tail Limits and Endpoint Atoms

A finite endpoint has a precise tail signature: $\overline F(x)$ is zero at and beyond the endpoint, but its limit from below need not be zero. If $X$ has an atom at $b$, that mass is still included in $P(X>x)$ for every $x<b$, however close $x$ is to $b$. At $x=b$, the strict tail excludes it.

Endpoint limit: If $b$ is a finite upper endpoint, then $$ \overline F(x)=0\quad\text{for }x\ge b, \qquad \lim_{x\uparrow b}\overline F(x)=P(X=b). $$ Thus the left-hand limit is zero when there is no atom at $b$, and equals the endpoint mass when there is one.

The one-sided limit follows directly from continuity of probability for decreasing events. As $x$ increases to $b$ from below, the events $\{X>x\}$ decrease to $\{X\ge b\}$. Their probabilities therefore tend to $P(X\ge b)$. Because $b$ is an upper endpoint, $P(X>b)=0$, so $P(X\ge b)=P(X=b)$.

Worked Example: A Finite Endpoint With an Atom

Suppose $X$ has the discrete distribution $$ P(X=0)=0.2,\qquad P(X=2)=0.5,\qquad P(X=5)=0.3. $$ The support is $\{0,2,5\}$, so its upper endpoint is $b=5$.

For $2\le x<5$, the only outcome greater than $x$ is $X=5$, so $\overline F(x)=0.3$. At $x=5$, no outcome is greater than $x$, and $\overline F(5)=0$. In particular, $$ \lim_{x\uparrow 5}\overline F(x)=0.3=P(X=5). $$

The CDF gives the same endpoint information. For $2\le x<5$, $F(x)=0.2+0.5=0.7$, while $F(x)=1$ for $x\ge5$. The jump from $0.7$ to $1$ at $5$ has size $0.3$, exactly the probability mass at the endpoint.

Unbounded Support: The Tail Vanishes but Never Reaches Zero

A basic probability limit explains why tail probabilities go to zero even for distributions with unbounded support. As $x$ increases without bound, the events $\{X>x\}$ decrease to the empty set: every outcome of a real-valued random variable is finite. Continuity of probability for decreasing events gives $$ \lim_{x\to\infty}\overline F(x)=0. $$ This limit holds whether the upper endpoint is finite or infinite. The distinction is whether the tail becomes exactly zero at some finite threshold.

Unbounded-upper-support criterion: The support is unbounded above if and only if $F(x)<1$, equivalently $\overline F(x)>0$, for every finite real $x$. In that case $\overline F(x)$ still tends to zero as $x\to\infty$.

To see the criterion, if the support is unbounded above, then for every finite $x$ there is a support point above $x$, and a neighborhood of that point has positive probability. Hence $P(X>x)>0$. Conversely, if $P(X>x)>0$ for every finite $x$, the distribution cannot be concentrated below any finite bound, so its support is unbounded above.

Worked Example: An Unbounded Distribution With a Vanishing Tail

Let $X$ have an exponential distribution with rate $2$. Its CDF and strict survival function are $$ F(x)= \begin{cases} 0, & x<0,\\ 1-e^{-2x}, & x\ge0, \end{cases} \qquad \overline F(x)= \begin{cases} 1, & x<0,\\ e^{-2x}, & x\ge0. \end{cases} $$

For each finite $x\ge0$, $e^{-2x}>0$, so $F(x)<1$. There is no finite $c$ with $F(c)=1$, and the support $[0,\infty)$ is unbounded above. Nevertheless, the tail tends to zero: for example, $\overline F(2)=e^{-4}$, which is small but positive, and $\lim_{x\to\infty}e^{-2x}=0$.

The contrast with the uniform example is exact. For the uniform distribution the tail is zero at every $x\ge3$; for this exponential distribution it is positive at every finite $x\ge0$. Both tails tend to zero as $x\to\infty$.

Lower Endpoints and the Left Tail

The same reasoning works at the lower end. A finite lower endpoint $a$ means that $X\ge a$ almost surely and the support has no points below $a$. In CDF terms, $F(x)=0$ for every $x<a$, while $F(x)>0$ for every $x>a$. At $x=a$, $F(a)=P(X=a)$, which may be zero or positive. If the support is unbounded below, then $F(x)>0$ for every finite $x$, although $F(x)$ still tends to zero as $x\to-\infty$.

Worked Example: Identifying a Lower Endpoint From the CDF

Let $Y$ be uniform on $[1,4]$, with CDF $F_Y(x)=0$ for $x<1$, $F_Y(x)=(x-1)/3$ for $1\le x<4$, and $F_Y(x)=1$ for $x\ge4$. For $x<1$, $F_Y(x)=0$, so $Y$ cannot fall below $1$. For $x>1$ and $x<4$, $(x-1)/3>0$; for $x\ge4$, the CDF is one. The support is $[1,4]$, and its lower endpoint is $a=1$. Since this distribution has no atom at $1$, $F_Y(1)=0$, illustrating why a finite lower endpoint need not have positive probability at the endpoint itself.

Common Mistake and Practical Use

Common mistake: “The tail probability tends to zero, so the support must be bounded above.” This reasoning confuses a limiting probability with an exact cutoff. The exponential example refutes it: $P(X>x)$ tends to zero, but remains positive for every finite $x$. The correction is to ask whether the CDF actually equals one at a finite argument. If it does, the distribution has a finite upper endpoint; if not, and the CDF remains below one at every finite argument, its upper support is unbounded.

This check helps distinguish a genuine boundary of possible values from a distribution whose extreme values are simply rare. It also prevents a common interpretive error when reading a plotted CDF: a curve that appears flat near one may still be strictly below one, and visual closeness is not proof of a finite endpoint.

Check Your Understanding

Use the definitions and endpoint criteria to answer the following.

A CDF satisfies $F(x)=1$ for every $x\ge 7$ and $F(x)<1$ for every $x<7$. What is the upper endpoint, and what is the survival probability at $x=7$?
Why does $\lim_{x\to\infty}\overline F(x)=0$ fail to show that the support is bounded above?
If a finite upper endpoint $b$ has an atom of mass $0.12$, what is $\lim_{x\uparrow b}\overline F(x)$? Explain why this need not equal $\overline F(b)$.
What condition on $F$ at finite arguments characterizes support that is unbounded above?
A distribution has $F(x)=0$ for every $x< -3$, but $F(-3)=0.2$. What does this say about its lower endpoint and the probability at that endpoint?

What You'll Learn