The 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.

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.


2 Answers 2


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.

  • $\begingroup$ can you provide me with a proof of that result or a link to one? $\endgroup$ Commented Dec 19, 2017 at 1:42

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