CDF Boundary Limits at Infinity

Why CDFs Have Boundary Limits

A cumulative distribution function records the probability that a random variable lies at or below a threshold. As the threshold moves far to the left, that event should become negligible; as it moves far to the right, it should include essentially all possible outcomes. The precise statements are

\lim_{x\to-\infty}F(x)=0 \qquad\text{and}\qquad \lim_{x\to+\infty}F(x)=1.

These are consequences of probability itself, not assumptions about whether the distribution is discrete or continuous, bounded or unbounded. The proof uses countable additivity through continuity of probability for nested events. We establish the needed continuity fact first, then apply it to threshold events.

Probability Continuity for Nested Events

Let \(A_1,A_2,\ldots\) be events in a probability space. The sequence is increasing if \(A_n\subseteq A_{n+1}\) for each \(n\). It is decreasing if \(A_{n+1}\subseteq A_n\) for each \(n\). Countable additivity implies the following two limits.

Theorem: If \(A_n\) is an increasing sequence of events, then

P\left(\bigcup_{n=1}^{\infty} A_n\right) =\lim_{n\to\infty}P(A_n).

If \(A_n\) is a decreasing sequence of events, then

P\left(\bigcap_{n=1}^{\infty} A_n\right) =\lim_{n\to\infty}P(A_n).

For an increasing sequence, define disjoint events \(D_1=A_1\) and \(D_k=A_k\setminus A_{k-1}\) for \(k\ge 2\). Their first \(n\) terms have union \(A_n\), and all the \(D_k\) together have union \(\bigcup_{k=1}^{\infty}A_k\). Countable additivity therefore gives

P(A_n)=\sum_{k=1}^{n}P(D_k) \quad\text{and}\quad P\left(\bigcup_{k=1}^{\infty}A_k\right) =\sum_{k=1}^{\infty}P(D_k) =\lim_{n\to\infty}\sum_{k=1}^{n}P(D_k).

This proves the increasing-sequence statement. For a decreasing sequence, the complements \(A_n^c\) form an increasing sequence, and \(\bigcup_n A_n^c=(\bigcap_n A_n)^c\). Applying the result just proved and using \(P(B^c)=1-P(B)\) yields

P\left(\bigcap_{n=1}^{\infty}A_n\right) =1-P\left(\bigcup_{n=1}^{\infty}A_n^c\right) =1-\lim_{n\to\infty}P(A_n^c) =\lim_{n\to\infty}P(A_n).

The final equality follows because \(P(A_n^c)=1-P(A_n)\). This argument applies to decreasing event sequences because probabilities are finite, being between zero and one. \(\square\)

The Boundary-Limit Theorem

Definition: For a real-valued random variable \(X\), its cumulative distribution function is \(F(x)=P(X\le x)\), for every real number \(x\).

The phrase real-valued matters: \(X\) takes a finite real number at every outcome under consideration. In particular, it does not take the value \(-\infty\) or \(+\infty\). No assumption that \(X\) has a density, or that its possible values are bounded, is needed.

Theorem: If \(X\) is a real-valued random variable with CDF \(F\), then

\lim_{x\to-\infty}F(x)=0 \qquad\text{and}\qquad \lim_{x\to+\infty}F(x)=1.

Proof. First consider the thresholds \(-1,-2,\ldots\) and define \(A_n=\{X\le -n\}\). These events decrease as \(n\) increases. Their intersection is empty: an outcome in every \(A_n\) would have to satisfy \(X\le -n\) for every positive integer \(n\), which no finite real value can satisfy. By continuity for decreasing events,

\lim_{n\to\infty}F(-n) =\lim_{n\to\infty}P(A_n) =P\left(\bigcap_{n=1}^{\infty}A_n\right) =P(\varnothing)=0.

Now let \(x\) tend to \(-\infty\) through arbitrary real values, not only negative integers. If \(x\le -n\), then \(\{X\le x\}\subseteq\{X\le -n\}\), so

0\le F(x)\le F(-n).

Given any \(\varepsilon>0\), choose \(n\) large enough that \(F(-n)<\varepsilon\). Every \(x\le -n\) then satisfies \(0\le F(x)<\varepsilon\), proving the left-hand limit.

For the right-hand limit, set \(B_n=\{X\le n\}\). These events increase, and their union is the whole sample space: every finite real value of \(X\) is at most some positive integer. Continuity for increasing events gives

\lim_{n\to\infty}F(n) =\lim_{n\to\infty}P(B_n) =P\left(\bigcup_{n=1}^{\infty}B_n\right) =P(\Omega)=1.

For arbitrary \(x\ge n\), the inclusion \(\{X\le n\}\subseteq\{X\le x\}\) gives \(F(x)\ge F(n)\), while \(F(x)\le 1\) because \(F(x)\) is a probability. Given \(\varepsilon>0\), choose \(n\) so that \(F(n)>1-\varepsilon\). Then every \(x\ge n\) satisfies \(1-\varepsilon<F(x)\le1\), proving the right-hand limit. \(\square\)

Worked Examples: Checking Boundary Behavior

Worked Example: A Finite Distribution

Suppose \(X\) takes the values \(-2,1,4\), with probabilities \(0.2,0.5,0.3\), respectively. Check the limits of its CDF.

The probabilities are valid because \(0.2+0.5+0.3=1\). If \(x<-2\), none of the possible values is at or below \(x\), so \(F(x)=0\). If \(x\ge4\), all three possible values are at or below \(x\), so \(F(x)=0.2+0.5+0.3=1\). Therefore \(F(x)\) is exactly zero sufficiently far to the left and exactly one sufficiently far to the right. In particular,

\lim_{x\to-\infty}F(x)=0,\qquad \lim_{x\to+\infty}F(x)=1.

A finite distribution reaches both limiting values beyond its smallest and largest possible values. A distribution with unbounded possible values need not reach one at any finite threshold.

Worked Example: A Geometric Distribution With Unbounded Support

Let \(X\) take nonnegative integer values with \(P(X=k)=(1-r)r^k\) for \(k=0,1,2,\ldots\), where \(0<r<1\). Find the boundary limits of its CDF.

For any integer \(n\ge0\), the finite geometric sum gives

P(X\le n)=\sum_{k=0}^{n}(1-r)r^k =(1-r)\frac{1-r^{n+1}}{1-r} =1-r^{n+1}.

The masses define a probability law because their infinite sum is

\sum_{k=0}^{\infty}(1-r)r^k =(1-r)\frac{1}{1-r}=1.

If \(x<0\), then \(F(x)=0\). If \(x\ge0\), let \(n=\lfloor x\rfloor\), the greatest integer not exceeding \(x\). Since \(X\) only takes integer values, \(F(x)=P(X\le n)=1-r^{n+1}\). As \(x\to+\infty\), \(n\to\infty\) and \(r^{n+1}\to0\), since \(0<r<1\). Thus \(F(x)\to1\). On the left, \(F(x)\) is already zero for every \(x<0\), so its limit there is zero.

Worked Example: An Exponential Distribution

Suppose \(X\) has density \(f(x)=2e^{-2x}\) for \(x\ge0\), and \(f(x)=0\) for \(x<0\). Check that the resulting CDF has the required limits.

For \(x<0\), the event \(X\le x\) has probability zero, so \(F(x)=0\). For \(x\ge0\), integrate the density over the portion of its support at or below \(x\):

F(x)=\int_0^x 2e^{-2t}\,dt =\left[-e^{-2t}\right]_{0}^{x} =1-e^{-2x}.

The density is normalized, since \(\int_0^\infty 2e^{-2t}\,dt=1\). For \(x\to-\infty\), all thresholds are eventually negative and \(F(x)=0\). For \(x\to+\infty\), \(e^{-2x}\to0\), so \(F(x)=1-e^{-2x}\to1\). Unlike the finite example, this CDF approaches one without equaling one at any finite \(x\).

Worked Example: Rejecting a Function With the Wrong Right Limit

Consider the proposed function \(G(x)=0\) for \(x<0\), \(G(x)=1/2\) for \(0\le x<1\), and \(G(x)=3/4\) for \(x\ge1\). Could it be the CDF of a real-valued random variable?

On the far left, \(G(x)=0\), so its left-hand limit is zero. On the far right, however, \(G(x)=3/4\) for every \(x\ge1\), giving

\lim_{x\to+\infty}G(x)=\frac34\ne1.

Consequently \(G\) cannot be a CDF. Its values never account for the full probability mass required by a probability distribution on the real line. This test rules it out without needing to construct or identify any candidate random variable.

Using the Limits Carefully

The boundary limits are universal checks: they apply to every CDF of a real-valued random variable, whether its distribution has a finite set of possible values or extends across an unbounded range. They are especially useful when a formula is given in pieces. Inspect the far-left and far-right pieces, and calculate their limits rather than relying on how the formula looks near its breakpoints.

Passing this check is necessary, but it is not enough to establish that a function is a CDF. A function might approach zero and one at the appropriate ends and still fail other required properties. For instance, the boundary limits alone do not verify that the function's values are compatible with probabilities as the threshold changes. The limits are one part of a validity check, not a complete certification.

Common pitfall: Do not assume the right-hand limit must be attained at a finite value. The geometric and exponential examples approach one as the threshold grows, but neither has a finite threshold beyond which its CDF is identically one. A limit of one means the values can be made arbitrarily close to one; it does not require equality at some finite \(x\).

A second point to keep in view is the real-valued assumption in the theorem. The proof that the far-left events have empty intersection uses the fact that \(X\) cannot equal \(-\infty\). For the real-valued random variables considered here, the boundary limits follow directly from countable additivity and the definition of the CDF.

Key takeaway: For every real-valued random variable, \(F(x)=P(X\le x)\) satisfies \(\lim_{x\to-\infty}F(x)=0\) and \(\lim_{x\to+\infty}F(x)=1\). These limits follow from continuity of probability for nested events and provide necessary checks for proposed CDFs.

Check Your Understanding

Answer each question by referring to the event probabilities involved.

For \(A_n=\{X\le -n\}\), explain why the events decrease and why their intersection is empty when \(X\) is real-valued.
For \(B_n=\{X\le n\}\), identify their union and state which probability-continuity result gives the limit of \(P(B_n)\).
A finite distribution assigns all its probability to values between \(-3\) and \(8\). What are its CDF values for \(x<-3\) and \(x\ge8\)?
A proposed function is zero for every \(x<0\) and equals \(0.9\) for every \(x\ge2\). Which boundary limit does it fail, and what conclusion follows?
Why does a CDF that approaches one as \(x\to+\infty\) not necessarily equal one at any finite threshold?