# Proving Slutsky's theorem

How do we go about proving the following part of Slutsky's theorem?

If $X_n \xrightarrow{d} X,\quad Y_n \xrightarrow{P} c$, then $X_nY_n \xrightarrow{d} Xc$ where $c$ is a degenerate random variable.

I tried using the following fact:

If $|X_n-Y_n| \xrightarrow{P} 0, \quad Y_n \xrightarrow{d} Y$, then $X_n \xrightarrow{d} Y$.

However, I could not arrive at a continuous transformation to use this fact.

I tried by lim sup and lim inf approach, directly from the definition of convergence in distribution:

$X_n \xrightarrow{d} X$, if $F_n(x) \rightarrow F(x)$ at all points of continuity of $F$, where $F_n$ and $F$ are the distribution functions of $X_n$ and $X$ respectively.

Is there any other equivalent characterisation that can help me proving Slutsky's theorem?

• can you teach me how to prove the fact you used? Dec 9, 2018 at 9:16

The fact you mention reads as follows: if $Z_n\to Z$ in distribution and $Z'_n\to 0$ in probability, then $Z_n+Z'_n\to Z$ in distribution.
We have $$X_nY_n=X_n(Y_n-c)+cX_n;$$ defining $Z_n:=cX_n$ and $Z'_n:=X_n(Y_n-c)$, we reach the wanted conclusion provided that we manage to show that $X_n(Y_n-c)\to 0$ in probability. But for a fixed $\varepsilon$, and each $R$ $$\mathbb P\{| X_n(Y_n-c)|\gt \varepsilon\}\leqslant\mathbb P\{|X_n|\gt R\}+\mathbb P\{|Y_n-c|\gt \varepsilon/R \}.$$ Choosing $R$ as a limiting point of the distribution function of $|X|$, we obtain from the convergence of $Y_n$ to $c$ in probability that $$\limsup_{n\to +\infty} \mathbb P\left\{| X_n(Y_n-c)|\gt \varepsilon\right\}\leqslant \mathbb P\{|X|\gt R\}.$$ Since $R$ can be chosen arbitrarily large, we are done.