Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-07
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Slutsky's theorem for real random variables

Statement

Let (Xn) and (Yn) be real random variables on one probability space, and let X be a real random variable (possibly on another space). If Xn⇒X and Yn→c in probability for c∈R, then Xn+Yn⇒X+c and XnYn⇒cX. If c≠0, define Qn=Xn/Yn on {Yn≠0} and give Qn any fixed value on {Yn=0}. Then Qn⇒X/c.

Facts & Assumptions

Given: (Xn) and (Yn) are on one probability space; Xn⇒X, Yn→c in probability, and the displayed quotient convention when c≠0.

[L1]

Distributional convergence is CDF convergence at continuity points (Convergence in distribution for real random variables).

[L2]

Probability convergence makes P(∣Yn−c∣>δ)→0 for each δ>0 (Convergence in probability).

Proof

technique · direct
1.1L1L2

For real Un,Vn on a common probability space, if Un⇒U and ∣Vn−Un∣→0 in probability, then Vn⇒U. Indeed, for every δ>0, with pn=P(∣Vn−Un∣>δ), FUn(t−δ)−pn≤FVn(t)≤FUn(t+δ)+pn. At a continuity point t of FU, take δ↓0 through values for which both t−δ and t+δ are continuity points. These values exist because a CDF has at most countably many jumps (for each positive integer k, there are at most k jumps larger than 1/k). First let n→∞ for each such δ, then let δ↓0; [L1] and [L2] give the assertion.

1.2L1

The CDF definition [L1] gives both affine operations needed below. First, Xn+a⇒X+a because FXn+a(t)=FXn(t−a). It also gives aXn⇒aX for every constant a: for a>0 use FaXn(t)=FXn(t/a); for a<0, use FaXn(t)=1−FXn((t/a)−) and squeeze the left limit between FXn(t/a−δ) and FXn(t/a), taking δ↓0 through continuity points t/a−δ; and for a=0 the claim is immediate. At continuity points of the transformed limit CDF, the corresponding point of FX is a continuity point.

1.3L1L2

The sequence (Xn) is bounded in probability: CDF convergence [L1] at two continuity points outside a sufficiently large interval makes lim sup⁡nP(∣Xn∣>M) arbitrarily small. Therefore P(∣Xn(Yn−c)∣>ε)≤P(∣Xn∣>M)+P(∣Yn−c∣>ε/M) shows Xn(Yn−c)→0 in probability. If c≠0, on ∣Yn−c∣<∣c∣/2, ∣Qn−Xn/c∣≤2∣c∣2∣Xn∣∣Yn−c∣, and the exceptional event {∣Yn−c∣≥∣c∣/2} contains {Yn=0} and has probability at most P(∣Yn−c∣>∣c∣/4)→0; the same boundedness argument gives Qn−Xn/c→0 in probability.

2.1step 1.1step 1.2L2

Addition follows from step 1.1 with Un=Xn+c and Vn=Xn+Yn: Un⇒X+c by step 1.2, while ∣Vn−Un∣=∣Yn−c∣→0 in probability by [L2].

3.1step 1.1step 1.2step 1.3∎

Apply step 1.1 to Un=cXn, Vn=XnYn, using step 1.2 and step 1.3, to obtain XnYn⇒cX. When c≠0, apply it again to Un=Xn/c, Vn=Qn, to obtain Qn⇒X/c.

Depends on

Used by

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Sources