Slutsky's theorem

Remarks

Proof idea: since the limit of $Y_n$ is a constant, the pair $(X_n, Y_n)$ converges in distribution to $(X, c)$. Each of the three maps $(x, y) \mapsto x + y$, $xy$ and $x/y$ is continuous at the points $(x, c)$, so the continuous mapping theorem applies. Example: let $X_1, X_2, \dots$ be i.i.d. with mean $\mu$ and variance $\sigma^2 \in (0, \infty)$, with sample mean $\bar{X}_n$ and sample standard deviation $S_n$. The central limit theorem gives $\sqrt{n}(\bar{X}_n - \mu)/\sigma \to N(0, 1)$, and the law of large numbers gives $S_n \to \sigma$ in probability; so $\sqrt{n}(\bar{X}_n - \mu)/S_n \to N(0, 1)$ in distribution. The limit $c$ must be a constant. Let $Z$ be standard normal and $X_n = Z$; the variables $Z$ and $-Z$ have the same law. With $Y_n = -Z$ one gets $X_n + Y_n = 0$, and with $Y_n = Z$ one gets $2Z$. So convergence in law of both terms does not determine the law of the limit of the sum. The theorem holds in the same way for random vectors, and for random matrices $Y_n$ that converge in probability to a constant matrix $C$. For the quotient $Y_n^{-1} X_n$ one needs $C$ to be invertible.

Going deeper

For a student

Slutsky's theorem is used constantly in asymptotic statistics, together with the central limit theorem and the delta method. It justifies the use of estimated standard errors in confidence intervals and test statistics such as the Student $t$ statistic for large samples. It follows from the continuous mapping theorem applied to the pair $(X_n, Y_n)$. Its key point is that joint convergence comes for free when one limit is constant.