Is Slutsky's theorem still valid when two sequences both converge to a non-degenerate random variable?

Let $\{X_n\}$, $\{Y_n\}$ be two sequences of scalar/vector/matrix random elements.

If $X_n$ converges in distribution to a random element $X$ and $Y_n$ converges in probability to a constant $c$, then \eqalign{ X_{n}+Y_{n}\ &{\xrightarrow {d}}\ X+c\\ X_{n}Y_{n}\ &{\xrightarrow {d}}\ cX\\ X_{n}/Y_{n}\ &{\xrightarrow {d}}\ X/c, } provided that $c$ is invertible, where ${\xrightarrow {d}}$ denotes convergence in distribution.

If both sequences in Slutsky's theorem both converge to a non-degenerate random variable, is the theorem still valid, and if not (could someone provide an example?), what are the extra conditions to make it valid?

Slutsky's theorem does not extend to two sequences converging in distributions to a random variable. If $Y_n$ converges in distribution to $Y$, $X_n+Y_n$ may well fail to converge or may converge to something else than $X+Y$.

For instance, if $Y_n=-X_n$ for all $n$'s, $X_n+Y_n$ does not converge to the difference of two rv's with same distribution as $X$.

Another counter-example is that, when the sequences $\{X_n\}$ and $\{Y_n\}$ are independent and both converging in distribution to a normal $\text{N}(0,1)$ variable, if one defines $X_1\sim\text{N}(0,1)$ and $X_2=-X_1$, then \eqalign{X_{n}\ &{\xrightarrow {d}}\ X_1\\Y_{n}\ &{\xrightarrow {d}}\ X_2\\X_{n}+Y_{n}\ &\not{{\xrightarrow {d}}}\ X_1+X_2=0} See the answer by Davide for more details on this example.

• For it to extend you need something more, like independence. Dec 16, 2015 at 9:34
• Am I right in thinking that if both sequences instead converge to a constant, Slutsky DOES still apply because a constant is a special (degenerate) case of a RV? Jan 6, 2016 at 3:12
• @half-pass: this is correct. Jan 6, 2016 at 5:43

Assume that $(X_0,Y_0)$ is a Gaussian centered vector whose covariance matrix is $\pmatrix{1&\rho\\ \rho&1}$ with $|\rho|\leqslant 1$. Define $X_n:=X_0$ and $Y_n:=Y_0$ for $n\geqslant 1$. Then $X_n\to X$ and $Y_n\to Y$, where $X$ and $Y$ are standard normal random variable. However, $X_n+Y_n$ is Gaussian, centered and its variance is $2+2\rho$. Since nothing is known about the distribution of $X+Y$, we cannot assert that $X_n+Y_n\to X+Y$ in distribution.

This examples shosw that we may have in general $X_n\to X$ and $Y_n\to Y$ in distribution, but if we do not have information about the distribution of $X+Y$, the convergence $X_n+Y_n\to X+Y$ may fail.

Of course, everything is fine if $(X_n,Y_n)\to (X,Y)$ in distribution (for example if $X_n$ is independent of $Y_n$ and $X$ of $Y$. In general, we can only assert that the sequence $(X_n+Y_n)_{n\geqslant 1}$ is tight (that is, for each positive $\varepsilon$, we can find $R$ such that $\sup_n\mathbb P\{|X_n+Y_n|\gt R\}\lt \varepsilon$). This implies that we may find an increasing sequence of integers $(n_k)_{k\geqslant 1}$ such that $(X_{n_k}+Y_{n_k})_{k\geqslant 1}$ converges in distribution to some random variable $Z$.

Proposition. There exists sequences of Gaussian random variables $(X_n)_{n\geqslant 1}$ and $(Y_n)_{n\geqslant 1}$ such that for any $\sigma\in [0,2]$, we can find an increasing sequence of integers $(n_k)_{k\geqslant 1}$ such that $(X_{n_k}+Y_{n_k})_{k\geqslant 1}$ converges in distribution to $N(0,\sigma^2)$.

Proof. Consider an enumeration $(r_j)$ of rational numbers of $[-1,1]$ and a bijection $\tau\colon \mathbb N\to \mathbb N^2$. For $n\in \tau^{-1}(\{j\})\times\mathbb N$, define $(X_n,Y_n)$ as a Gaussian centered vector of covariance matrix $\pmatrix{1&r_j\\ r_j&1}$. With this choice, one can see that the conclusion of the proposition is satisfied when $\sigma$ is rational. Use an approximation argument for the general case.