# How does Slutsky's theorem extends when two random variables converge to two constants?

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.

Now suppose that $X_n$ also converges in probability to a constant $a$. Can I conclude that: $$X_{n}Y_{n}\ {\xrightarrow {p}}\ ac$$

If so, on what grounds? Does this need a separate proof, or does it in some way - which I do not see - follow trivially from the theorem?

Note that there is already a good answer for when $Y_n$ converges in distribution to Y, that the theorem does not extend to such cases.

Convergence in probability is stronger than convergence in distribution, and thus, you can conclude that given $X_n \overset{p}{\to}a$, $X_n \overset{d}{\to}a$. You can find a proof of that fact here. Thus, Slutsky's theorem applies directly, and

$$X_n Y_n \overset{d}{\to} ac.$$

Now, when a random variable $Z_n$ converges in distribution to a constant, then it also converges in probability to a constant, you can find the proof here.

Thus, $X_n Y_n \overset{p}{\to} ac$.

Yes, your result is true, as explained by Greenparker. In fact, a much stronger result holds, which does not follow from Slutsky. It even holds that if $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$ (i.e. convergence to random variables rather than constants), then you still have:

$$X_n Y_n \overset{p}{\to} XY$$

To prove this you use the following standard result, which you will find in any probability textbook: If $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$, then:

$$(X_n, Y_n) \overset{p}{\to} (X,Y)$$

Now use the continuous mapping theorem with $\phi(x,y)= xy$ to get the stated result.

• can you provide me with a proof of that result or a link to one? – Jeff Dec 19 '17 at 1:42