Convergence in Probability¶

Definition¶

Let $X_1,X_2,\ldots$ and $X$ be random variables on a common probability space. The sequence converges in probability to $X$ , written $X_n\xrightarrow{P}X$ $X$ , if for every $\varepsilon>0$ $0$ ,

\mathbb{P}(|X_n-X|>\varepsilon)\longrightarrow0\quad\text{as }n\to\infty.

The Sequence consists of random variables $X_n$ . For each fixed positive tolerance, the Limit of a Sequence is instead taken for the numbers $\mathbb{P}(|X_n-X|>\varepsilon)$ $ε)$ , which must tend to zero.

Interpretation¶

Convergence in probability asks, for each fixed tolerance $\varepsilon>0$ $0$ , whether the probability of an error larger than $\varepsilon$ tends to zero. It makes a statement about each sample size $n$ separately, not about how a single sample path behaves as $n$ grows.

Two densities of X n centred at c, a wide one for small n and a narrow one for large n, with a band from c minus epsilon to c plus epsilon; the shaded tails outside the band shrink as n grows.

The figure shows the density of $X_n$ for a small and a large $n$ , both centred at $c$ , with the band from $c-\varepsilon$ to $c+\varepsilon$ marked. The shaded area outside the band is $\mathbb{P}(|X_n-c|>\varepsilon)$ $ε)$ , and it is much smaller for the narrow large- $n$ density. Convergence in probability to $c$ means that this area tends to $0$ as $n\to\infty$ $\infty$ , however narrow the band is chosen.

Two steps are involved: first $\varepsilon$ is fixed, then the limit of the numbers $\mathbb{P}(|X_n-X|>\varepsilon)$ is taken. Small errors are allowed at every $n$ ; only the probability of errors above the tolerance must vanish. Most uses have a constant limit, as in the weak law of large numbers. A quick proof for a constant limit uses the Chebyshev inequality: if $\mathbb{E}X_n=c$ and $\operatorname{Var}(X_n)\to0$ $0$ , then $\mathbb{P}(|X_n-c|\ge\varepsilon)\le\operatorname{Var}(X_n)/\varepsilon^2\to0$ $0$ .

A common slip is to conclude that expectations converge as well. If $X_n=n$ with probability $1/n$ and $X_n=0$ otherwise, then $X_n\xrightarrow{P}0$ $0$ , yet $\mathbb{E}X_n=1$ for every $n$ : a rare but large error does not count in the definition. The stronger almost-sure convergence additionally requires almost every path to converge.

Key Results¶

Almost-sure convergence implies convergence in probability.
Convergence in probability implies convergence in distribution; the converses generally fail.

Examples¶

If $X_n=1/n$ $1/ n$ , then for every $\varepsilon>0$ $0$ , $\mathbb{P}(|X_n|>\varepsilon)=0$ for sufficiently large $n$ , so $X_n\xrightarrow{P}0$ $0$ .
If $X_n\sim\operatorname{Bern}(1/n)$ $Bern (1/ n)$ , then for $0<\varepsilon<1$ $1$ , $\mathbb{P}(|X_n|>\varepsilon)=1/n\to0$ $0$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 7

Keywords: convergence in probability to a constant, stochastic convergence Symbols: epsilon, infinity

Last updated: 20 October 2026 03:02