Why Countably Many Tolerances Can Be Enough
Convergence in probability is defined by a condition for every positive tolerance: the probability that the error exceeds that tolerance must tend to zero. At first glance, this requires checking uncountably many limits. In practice, a countable collection can suffice. The reason is geometric: between any positive tolerance and zero lie positive rational numbers, and a smaller tolerance gives a larger error event.
We make this precise for real-valued random variables \(X_n\) and \(X\) defined on the same probability space. For each \(n\) and each \(t>0\), write \(A_n(t)=\{|X_n-X|>t\}\). These events are measurable because \(X_n\) and \(X\) are random variables and the absolute-value map is measurable.
For fixed \(n\), the events are nested as the threshold changes. If \(0<s<t\), then $$ A_n(t)\subseteq A_n(s), $$ because an error greater than \(t\) is also greater than \(s\). Consequently, $$ P(A_n(t))\leq P(A_n(s)). $$ This simple inclusion is the key: to control the probability at an arbitrary tolerance, it is enough to control it at a smaller tolerance.
The Rational-Tolerance Criterion
The forward implication follows immediately from the definition, since every positive rational is a positive real number. The reverse implication is the useful part.
Let \(\varepsilon>0\). The rational numbers are dense in the real numbers, so choose a positive rational \(q\) with \(q<\varepsilon\).
If \(|X_n-X|>\varepsilon\), then \(|X_n-X|>q\). Thus \(\{|X_n-X|>\varepsilon\}\subseteq\{|X_n-X|>q\}\).
The probability at \(\varepsilon\) is bounded above by the probability at \(q\), which tends to zero by assumption.
In full, for the chosen \(q\), $$ 0\leq P(|X_n-X|>\varepsilon) \leq P(|X_n-X|>q)\longrightarrow 0. $$ Therefore \(P(|X_n-X|>\varepsilon)\to0\). Since \(\varepsilon>0\) was arbitrary, the definition gives \(X_n\xrightarrow{p}X\). \(\square\)
The argument uses only density of the positive rationals and event inclusion. No independence assumption is needed, and the distribution of \(X_n-X\) may change with \(n\). The theorem does not say that a single rational tolerance is enough: it requires convergence of the error probability at every positive rational tolerance.
Nested Thresholds and a Smaller Test Set
The same reasoning gives a more flexible criterion. Instead of checking every positive rational, one may check any fixed sequence of positive thresholds that decreases to zero. The thresholds need not themselves be rational.
If \(X_n\xrightarrow{p}X\), the stated limits hold by the definition. Conversely, fix \(\varepsilon>0\). Since \(r_k\to0\), there is some fixed \(k\) for which \(r_k<\varepsilon\). For this \(k\), event inclusion gives $$ 0\leq P(|X_n-X|>\varepsilon) \leq P(|X_n-X|>r_k)\longrightarrow 0. $$ This proves the definition for the arbitrary \(\varepsilon\), and hence proves the criterion. \(\square\)
For example, \(r_k=1/k\) is a convenient choice. At each fixed \(n\), the events \(\{|X_n-X|>1/k\}\) grow as \(k\) increases: a smaller threshold admits more errors. They are nested across thresholds, not necessarily across sample sizes \(n\). The theorem requires taking the limit in \(n\) separately for each fixed \(k\); it does not allow \(k\) to vary with \(n\) without additional argument.
Worked Applications
Worked Example: A Deterministic Sequence
Let \(X_n=1/n\) with probability one, and let \(X=0\). Verify convergence in probability using positive rational tolerances.
Fix a positive rational \(q\). Since the variables are deterministic, $$ P(|X_n-X|>q)=P(1/n>q). $$ Choose an integer \(N\) with \(N\geq 1/q\). For every \(n\geq N\), \(1/n\leq1/N\leq q\), so the strict inequality \(1/n>q\) is false. The probability is therefore zero for all \(n\geq N\), and its limit is zero. This holds for every positive rational \(q\); the rational-tolerance criterion gives \(X_n\xrightarrow{p}0\).
Worked Example: Rare Errors of Fixed Size
Let \(X=0\), and define \(X_n\) to equal \(1\) with probability \(1/n\) and \(0\) with probability \(1-1/n\). Check the criterion using the thresholds \(r_k=1/k\).
Fix \(k\). If \(k=1\), the event \(\{|X_n|>1\}\) is empty, so its probability is zero. If \(k\geq2\), then \(0<1/k<1\). The event \(\{|X_n|>1/k\}\) occurs exactly when \(X_n=1\), and hence $$ P(|X_n|>1/k)=\frac{1}{n}\longrightarrow0. $$ Thus the required limit holds for every fixed \(k\). Since \(1/k\to0\), the nested-threshold criterion gives \(X_n\xrightarrow{p}0\). The values of \(X_n\) do not get smaller on the rare event; convergence follows because the probability of that event vanishes.
Worked Example: An Unbounded Random Variable Divided by \(n\)
Let \(Y\) have the geometric distribution \(P(Y=m)=2^{-m}\) for \(m=1,2,\ldots\), and set \(X_n=Y/n\) and \(X=0\). Verify convergence using \(r_k=1/k\).
The probabilities defining \(Y\) sum to one because \(\sum_{m=1}^{\infty}2^{-m}=1\). For any \(t\geq0\), the first integer strictly greater than \(t\) is \(\lfloor t\rfloor+1\), so $$ P(Y>t)=\sum_{m=\lfloor t\rfloor+1}^{\infty}2^{-m} =2^{-\lfloor t\rfloor}. $$ For fixed \(k\), $$ P(|X_n|>1/k)=P(Y>n/k)=2^{-\lfloor n/k\rfloor}. $$ As \(n\to\infty\) with \(k\) fixed, \(\lfloor n/k\rfloor\to\infty\), so this probability tends to zero. The criterion yields \(X_n\xrightarrow{p}0\). In particular, the argument does not require \(Y\) to be bounded: its geometric tail makes the probability of exceeding the growing threshold \(n/k\) vanish.
Common Mistake and Why the Criterion Is Useful
A frequent error is to choose a rational \(q>\varepsilon\) and use the event inclusion in the wrong direction. If \(q>\varepsilon\), then $$ \{|X_n-X|>q\}\subseteq\{|X_n-X|>\varepsilon\}. $$ A probability tending to zero for the smaller event on the left does not bound the probability of the larger event on the right from above. To prove convergence at \(\varepsilon\), choose \(q<\varepsilon\), so that the \(\varepsilon\)-event is contained in the \(q\)-event.
This criterion is useful when a proof naturally produces bounds at a countable set of thresholds, or when estimates are easiest at thresholds such as \(1/k\). It preserves the full definition of convergence in probability while avoiding a separate argument for every positive real number. The essential logic is always the same: make the checked threshold smaller than the tolerance of interest, then use containment of the corresponding error events.
Check Your Understanding
Use event inclusion and keep the threshold fixed while taking each limit in \(n\).
- Suppose the error probability tends to zero at every positive rational threshold. For a given \(\varepsilon>0\), which direction should the rational threshold \(q\) lie relative to \(\varepsilon\), and why?
- Let \(r_k=2^{-k}\). Explain why convergence of \(P(|X_n-X|>r_k)\) to zero for every fixed \(k\) implies convergence in probability.
- If \(0<s<t\), which of the events \(\{|X_n-X|>s\}\) and \(\{|X_n-X|>t\}\) contains the other?
- For the rare-error example, what is \(P(|X_n|>1/2)\), and what is its limit as \(n\to\infty\)?
- Why does a limit established only at thresholds \(q>\varepsilon\) fail to provide the needed upper bound at \(\varepsilon\)?