A Truncated-Distance Criterion for Convergence in Probability

Why Truncate the Distance?

Convergence in probability asks whether large errors become unlikely: for every fixed tolerance, the probability that the error exceeds it must tend to zero. An expectation of the ordinary absolute error can be a poor summary of that behavior. A very large error on a very rare event may have little effect on the probability criterion but still dominate the ordinary expectation.

Truncating the error at one resolves this mismatch. The quantity $\min(1,|X_n-X|)$ records small errors at their actual size and assigns the same maximum value to every error of size at least one. Its expectation therefore responds to how often errors occur without allowing extremely large errors to dominate.

Definition: Let $X_n$ and $X$ be real-valued random variables on the same probability space. The truncated-distance expectation between $X_n$ and $X$ is $$ E[\min(1,|X_n-X|)]. $$ The integrand is measurable and bounded between zero and one, so this expectation is finite whether or not $X_n-X$ is integrable.

The Truncated-Distance Criterion

Recall that $X_n$ converges in probability to $X$ when, for every $\varepsilon>0$, $P(|X_n-X|>\varepsilon)\to 0$. The truncated-distance criterion replaces this family of probability limits with a single sequence of expectations.

Theorem: For real-valued random variables $X_n$ and $X$ on the same probability space, $$ X_n\xrightarrow{p}X \quad\Longleftrightarrow\quad E[\min(1,|X_n-X|)]\longrightarrow 0. $$ No moment or integrability assumption on $X_n-X$ is required.

Proof. Write $D_n=\min(1,|X_n-X|)$. First suppose $X_n\xrightarrow{p}X$. Fix any $\eta$ with $0<\eta<1$. On the event $\{|X_n-X|\leq\eta\}$, $D_n\leq\eta$; on its complement, $D_n\leq1$. Thus

D_n\leq \eta+\mathbf{1}_{\{|X_n-X|>\eta\}}, \qquad E[D_n]\leq \eta+P(|X_n-X|>\eta).

For this fixed $\eta$, convergence in probability makes the probability on the right tend to zero. Therefore $\limsup_{n\to\infty}E[D_n]\leq\eta$. This holds for every $\eta\in(0,1)$, and $E[D_n]\geq0$, so $E[D_n]\to0$.

Conversely, fix any $\varepsilon>0$. On $\{|X_n-X|>\varepsilon\}$, the truncated error is at least $\min(1,\varepsilon)$. Consequently,

\min(1,\varepsilon)\mathbf{1}_{\{|X_n-X|>\varepsilon\}} \leq D_n, \qquad P(|X_n-X|>\varepsilon) \leq \frac{E[D_n]}{\min(1,\varepsilon)}.

The denominator is positive and fixed as $n$ varies. If $E[D_n]\to0$, the probability tends to zero for every $\varepsilon>0$, which is precisely convergence in probability. $\square$

Worked Examples

Worked Example: A Deterministic Error

Let $X=0$ and let $X_n=1/n$ with probability one. Compute the truncated-distance expectation and verify convergence in probability.

The error is the constant $|X_n-X|=1/n$. Hence

E[\min(1,|X_n-X|)] =\min(1,1/n) =1/n \longrightarrow 0.

Directly, for any fixed $\varepsilon>0$, choose $n>1/\varepsilon$. Then $1/n<\varepsilon$, so $P(|X_n-X|>\varepsilon)=0$ for every such $n$. Thus $X_n\xrightarrow{p}X$, in agreement with the criterion.

Worked Example: Bounded Random Errors

Take $\Omega=[0,1]$ with the uniform probability distribution, set $A_n=[0,1/n]$, let $X=0$, and define $X_n=\mathbf{1}_{A_n}$. Find the truncated-distance expectation and check the probability criterion.

The errors are either zero or one, so they are bounded by one and truncation does not change them:

E[\min(1,|X_n-X|)] =E[\mathbf{1}_{A_n}] =P(A_n) =\frac{1}{n} \longrightarrow 0.

If $0<\varepsilon<1$, then $\{|X_n-X|>\varepsilon\}=A_n$, with probability $1/n$. If $\varepsilon\geq1$, the event is empty. In either case its probability tends to zero. Here the ordinary mean absolute error also tends to zero, since $E[|X_n-X|]=1/n$.

Worked Example: Rare Errors With Unbounded Magnitude

Again use $[0,1]$ with the uniform probability distribution, and set $A_n=[0,1/n]$. Let $X=0$ and $X_n=n^2\mathbf{1}_{A_n}$. These errors are not bounded by a common constant. Calculate both the truncated and ordinary mean absolute errors.

On $A_n$, the error has size $n^2$, so its truncated size is $\min(1,n^2)=1$. Off $A_n$, both sizes are zero. Therefore

E[\min(1,|X_n-X|)] =P(A_n)\cdot 1 =\frac{1}{n} \longrightarrow 0.

For any fixed $\varepsilon>0$, once $n^2>\varepsilon$, the event $\{|X_n-X|>\varepsilon\}$ is exactly $A_n$. Its probability is $1/n$, so $X_n\xrightarrow{p}X$. But the ordinary mean absolute error is

E[|X_n-X|] =n^2P(A_n) =n^2\frac{1}{n} =n,

which grows rather than tending to zero. The rare event has probability $1/n$, but its error size $n^2$ makes its contribution to the ordinary expectation large. Truncation preserves the conclusion about convergence in probability without requiring the ordinary mean absolute error to vanish.

Why the Criterion Is Useful

The theorem separates the probability of an error from its magnitude. Errors below one retain their actual size; errors at least one contribute one, whether their magnitude is two or two million. This is exactly what is needed for convergence in probability, which is about the chance of exceeding each fixed tolerance rather than about the total size of all errors.

The criterion also gives a natural distance between random variables. Define $\rho(X,Y)=E[\min(1,|X-Y|)]$. This is a metric on equivalence classes of real-valued random variables, where variables equal almost surely are treated as the same class.

Result: The function $\rho(X,Y)=E[\min(1,|X-Y|)]$ is nonnegative and symmetric, vanishes exactly when $X=Y$ almost surely, and satisfies the triangle inequality. Convergence in this metric is exactly convergence in probability.

Verification. Nonnegativity and symmetry follow from the corresponding properties of absolute value. If $\rho(X,Y)=0$, the nonnegative random variable $\min(1,|X-Y|)$ has expectation zero and is therefore zero almost surely. This means $X=Y$ almost surely. Conversely, almost-sure equality makes $\rho(X,Y)=0$.

For the triangle inequality, the absolute-value inequality gives $|X-Z|\leq|X-Y|+|Y-Z|$. For nonnegative $a,b$, $\min(1,a+b)\leq\min(1,a)+\min(1,b)$: if either $a$ or $b$ is at least one, the right side is at least one; otherwise it equals $a+b$, which is at least the left side. Applying these inequalities pointwise and taking expectations gives

\rho(X,Z) \leq E[\min(1,|X-Y|)]+E[\min(1,|Y-Z|)] =\rho(X,Y)+\rho(Y,Z).

The theorem proves that $\rho(X_n,X)\to0$ holds exactly when $X_n\xrightarrow{p}X$. Thus the truncated expectation does not merely provide a convenient test; it supplies a distance that captures this mode of convergence. $\square$

Common Pitfall: Requiring the Mean Absolute Error to Vanish

A common mistake is to replace the truncated-distance criterion with $E[|X_n-X|]\to0$ and treat that as the definition or necessary condition for convergence in probability. The rare-error example shows why this fails: there, convergence in probability holds even though the mean absolute error equals $n$.

If the errors are bounded by one, then truncation does nothing and the two expectations are equal. More generally, if $|X_n-X|\leq M$ almost surely for a fixed $M\geq1$, then

\min(1,|X_n-X|) \leq |X_n-X| \leq M\min(1,|X_n-X|).

So under that uniform bound, the truncated expectation tends to zero exactly when the mean absolute error does. Without such a bound, the truncated criterion remains valid, whereas the ordinary expectation can be large or may not exist. The correction is to match the condition being checked to the mode of convergence: use the truncated expectation for convergence in probability.

Check Your Understanding

Use the theorem and its proof to answer these questions.

Why is $E[\min(1,|X_n-X|)]$ finite even if $E[|X_n-X|]$ is infinite?
In the forward direction of the proof, why is it important to choose $0<\eta<1$?
If $E[\min(1,|X_n-X|)]\to0$, what upper bound does the proof give for $P(|X_n-X|>\varepsilon)$?
For an error equal to $n^3$ on an event of probability $1/n^2$ and zero otherwise, calculate its truncated-distance expectation for $n\geq1$. Does the criterion imply convergence in probability to zero?
Why does the rare-error example show that convergence in probability need not imply convergence of the mean absolute error?