Convergence in Distribution to a Constant

Why a Constant Limit Is Special

Convergence in distribution and convergence in probability are distinct notions in general. A useful exception occurs when the proposed limiting distribution is concentrated at a single number \(c\). In that case, the distribution function is especially simple: it is zero to the left of \(c\), one at and to the right of \(c\), and has its only jump at \(c\). Convergence of the distribution functions at their continuity points then forces the random variables to be close to \(c\) with high probability.

The result has two directions. Convergence in probability implies convergence in distribution, as established by the standard implication between these modes. Here we prove both directions for a constant limit, including why the converse works and why the jump point itself must be handled carefully. Assume throughout that \(X_n\) and the constant \(c\) are defined on a common probability space.

Definitions and the Main Equivalence

Definition: The point mass at \(c\), denoted \(\delta_c\), is the distribution assigning probability one to the value \(c\). Its CDF is \(F_{\delta_c}(x)=0\) for \(x<c\) and \(F_{\delta_c}(x)=1\) for \(x\ge c\). A sequence \(X_n\) converges in distribution to \(\delta_c\), written \(X_n\xrightarrow{d}\delta_c\), if \(F_n(x)\to F_{\delta_c}(x)\) at every continuity point \(x\) of \(F_{\delta_c}\). A sequence converges in probability to \(c\), written \(X_n\xrightarrow{p}c\), if for every \(\varepsilon>0\), \(P(|X_n-c|>\varepsilon)\to0\).

The CDF \(F_{\delta_c}\) is discontinuous at \(c\) and continuous at every \(x\ne c\). Thus the definition of convergence in distribution to \(\delta_c\) requires convergence of \(F_n(x)\) for all \(x<c\) and all \(x>c\), but it imposes no condition at \(x=c\).

Theorem: For random variables \(X_n\) on a common probability space and a constant \(c\), \(X_n\xrightarrow{d}\delta_c\) if and only if \(X_n\xrightarrow{p}c\).

From Distributional Convergence to Probability Convergence

Suppose \(X_n\xrightarrow{d}\delta_c\), and fix any \(\varepsilon>0\). The event that \(X_n\) is more than \(\varepsilon\) away from \(c\) splits into a left tail and a right tail:

\{|X_n-c|>\varepsilon\} = \{X_n<c-\varepsilon\}\,\cup\,\{X_n>c+\varepsilon\}.

These events are disjoint. The probability of the left tail is at most \(F_n(c-\varepsilon)\), and the probability of the right tail is \(1-F_n(c+\varepsilon)\). Both \(c-\varepsilon\) and \(c+\varepsilon\) are continuity points of \(F_{\delta_c}\). By distributional convergence, \(F_n(c-\varepsilon)\to0\) and \(F_n(c+\varepsilon)\to1\). Consequently,

\begin{aligned} 0 &\le P(|X_n-c|>\varepsilon)\\ &\le F_n(c-\varepsilon)+1-F_n(c+\varepsilon) \longrightarrow 0+1-1=0. \end{aligned}

Since this holds for every \(\varepsilon>0\), the epsilon-event definition gives \(X_n\xrightarrow{p}c\). This argument also shows the practical force of CDF convergence on both sides of the proposed constant: neither tail can retain positive probability.

From Probability Convergence to Distributional Convergence

Now suppose \(X_n\xrightarrow{p}c\). To prove convergence in distribution, take an arbitrary continuity point \(x\) of \(F_{\delta_c}\). Such an \(x\) must satisfy either \(x<c\) or \(x>c\). We verify CDF convergence in these two cases.

If \(x<c\), set \(\eta=(c-x)/2>0\). Whenever \(X_n\le x\), we have \(|X_n-c|\ge c-x>\eta\). Therefore,

0\le F_n(x)=P(X_n\le x) \le P(|X_n-c|>\eta)\longrightarrow0 =F_{\delta_c}(x).

If \(x>c\), set \(\eta=(x-c)/2>0\). Whenever \(X_n>x\), we have \(|X_n-c|>x-c>\eta\). Hence,

\begin{aligned} 0\le 1-F_n(x) &=P(X_n>x)\\ &\le P(|X_n-c|>\eta)\longrightarrow0. \end{aligned}

It follows that \(F_n(x)\to1=F_{\delta_c}(x)\) for \(x>c\). We have proved convergence at every continuity point of \(F_{\delta_c}\), so \(X_n\xrightarrow{d}\delta_c\). Together, the two arguments prove the theorem. \(\square\)

Worked Examples

Worked Example: A Deterministic Sequence Approaching c

Let \(X_n=c+3/n\) with probability one. Show directly that it converges both in probability and in distribution to the point mass at \(c\).

For any fixed \(\varepsilon>0\), choose an integer \(N>3/\varepsilon\). For every \(n\ge N\), \(3/n<\varepsilon\), so

P(|X_n-c|>\varepsilon) = \begin{cases} 1,&3/n>\varepsilon,\\ 0,&3/n\le\varepsilon \end{cases} =0\quad\text{for all }n\ge N.

Thus the probabilities tend to zero and \(X_n\xrightarrow{p}c\); the theorem gives \(X_n\xrightarrow{d}\delta_c\). Directly, if \(x<c\), then \(X_n>x\) for every \(n\), so \(F_n(x)=0\). If \(x>c\), then \(c+3/n\le x\) for all sufficiently large \(n\), so \(F_n(x)=1\) eventually. These are precisely the required CDF limits. No condition is required at \(x=c\).

Worked Example: Rare Outliers That Move Far Away

For each positive integer \(n\), let \(X_n=c+n\) with probability \(1/(n+1)\), and let \(X_n=c\) with probability \(n/(n+1)\). Verify convergence in probability, even though the possible outlier’s distance from \(c\) grows.

Fix \(\varepsilon>0\). For every integer \(n>\varepsilon\), the value \(c+n\) is more than \(\varepsilon\) from \(c\), while the value \(c\) is not. Therefore,

P(|X_n-c|>\varepsilon)=\frac{1}{n+1}\longrightarrow0.

This verifies \(X_n\xrightarrow{p}c\), and hence \(X_n\xrightarrow{d}\delta_c\). The example illustrates that convergence to a constant does not require every possible value of \(X_n\) to be close to \(c\). It requires the probability of being far away to vanish. Here the outlier becomes more distant, but its probability tends to zero.

Worked Example: A Sequence That Does Not Converge to a Constant

Let \(X_n\) take the values \(c-1\) and \(c+1\), each with probability \(1/2\), for every \(n\). Check whether this sequence converges in probability or distribution to \(c\).

With \(\varepsilon=1/2\), both possible values are more than \(\varepsilon\) from \(c\), so

$$ P(|X_n-c|>1/2)=1 $$

for every \(n\). The sequence does not converge in probability to \(c\). To check distributional convergence, take the continuity point \(x=c-1/2<c\). The value \(c-1\) is at most \(x\), while \(c+1\) is greater than \(x\), so \(F_n(x)=1/2\) for every \(n\). But \(F_{\delta_c}(x)=0\). The CDFs fail to converge at this continuity point, so \(X_n\) does not converge in distribution to \(\delta_c\) either.

A Useful Consequence and a Common Pitfall

The equivalence also ensures that a sequence cannot converge in probability to two different constants. Indeed, suppose \(X_n\xrightarrow{p}c\) and \(X_n\xrightarrow{p}d\). If \(c\ne d\), let \(\varepsilon=|c-d|/3\). The events \(\{|X_n-c|\le\varepsilon\}\) and \(\{|X_n-d|\le\varepsilon\}\) are disjoint: a value belonging to both would imply \(|c-d|\le2\varepsilon<|c-d|\) by the triangle inequality. But convergence in probability to each constant makes the probability of each event tend to one. Two disjoint events cannot both have probabilities tending to one, since their probabilities sum to at most one. Therefore \(c=d\).

Common pitfall: checking the CDF at \(c\). Since \(F_{\delta_c}\) jumps at \(c\), that point is not a continuity point. The definition of convergence in distribution does not require \(F_n(c)\to F_{\delta_c}(c)\). For example, in the rare-outlier construction above, \(F_n(c)=n/(n+1)\to1\), but that value is not needed to establish distributional convergence. The correct argument checks points strictly below and strictly above \(c\), or uses the theorem to establish convergence in probability.

Key takeaway: A point mass has no spread left to distinguish the two modes of convergence: \(X_n\xrightarrow{d}\delta_c\) holds exactly when \(X_n\xrightarrow{p}c\). The CDF criterion tests the two sides of \(c\), while the probability criterion directly controls the chance of being farther than any fixed tolerance.

Check Your Understanding

Use the definitions and the two directions of the theorem to answer these questions.

At which real numbers is the CDF of \(\delta_c\) continuous, and why does the convergence-in-distribution definition omit \(c\)?
If \(X_n\xrightarrow{d}\delta_c\), which two probabilities bound \(P(|X_n-c|>\varepsilon)\)?
In proving the reverse implication at a point \(x<c\), why is it useful to choose \(\eta=(c-x)/2\) rather than \(\eta=c-x\)?
Suppose \(P(X_n=c)=1-1/n\) and \(P(X_n=c+10)=1/n\) for \(n\ge2\). Does \(X_n\) converge in probability to \(c\)? Explain using an arbitrary fixed \(\varepsilon>0\).
For the sequence that assigns probability \(1/2\) to each of \(c-1\) and \(c+1\), give a continuity point of \(F_{\delta_c}\) where CDF convergence fails.