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I am confused about some details about Slutsky's theorem:

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?

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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.

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    $\begingroup$ For it to extend you need something more, like independence. $\endgroup$ – kjetil b halvorsen Dec 16 '15 at 9:34
  • $\begingroup$ 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? $\endgroup$ – half-pass Jan 6 '16 at 3:12
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    $\begingroup$ @half-pass: this is correct. $\endgroup$ – Xi'an Jan 6 '16 at 5:43
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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.

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