Slutsky's theorem and convergence in distribution

Question

I came across the following question, and I can't figure it out. Why does the counterexample work and why slutsky's theorem doesn't apply.

Slutsky's theorem for central limit theorem

$$ \begin{aligned} &\text { Assume } \sqrt{n}\left(X_{n}-X\right) \stackrel{d}{\rightarrow} N\left(0, \sigma^{2}\right) \text { and } Y_{n} \stackrel{p}{\rightarrow} c, \text { where } X \text { is a }\\ &\text { random variable and } c \text { is a positive constant. Does it hold that }\\ &\sqrt{n}\left(Y_{n} X_{n}-c X\right) \stackrel{d}{\rightarrow} N\left(0, c^{2} \sigma^{2}\right) ? \end{aligned} $$

As each XnYn aren't identically distributed, I tried applying the Lindeberg condition, but nothing is said about the squared convergence, so I reached an impasse.

thanks in advance

Answer 1

It might be worth quoting what Michael's counterexample was

No. Let $c=1, X=1,Y_n=1+1/n^{1/4}$

and presumably $X_n \sim N\left(1,\frac{\sigma^2}n\right)$ or something similar so $\sqrt{n}\left(X_{n}-X\right) \stackrel{d}{\rightarrow} N\left(0, \sigma^{2}\right)$.

If so, you would then get

• $Y_{n} X_{n} \sim N\left(1+\frac{1}{n^{1/4}},\left(1+\frac{1}{n^{1/4}}\right)^2\frac{\sigma^2}n\right) $
• $Y_{n} X_{n}-cX \sim N\left(\frac{1}{n^{1/4}},\left(1+\frac{1}{n^{1/4}}\right)^2\frac{\sigma^2}n\right) $
• $\sqrt{n}\left(Y_{n} X_{n}-c X\right) \sim N\left({n^{1/4}},\left(1+\frac{1}{n^{1/4}}\right)^2\sigma^2\right) $

and as $n$ increases, the variance in the final expression heads towards $\sigma^2$ as hoped for but the expectation slowly increases without limit and is eventually well away from $0$, all due to the slowness of $Y_n$'s convergence towards $c$.

Slutsky's theorem does not apply because this does not meet the preconditions for Slutsky's theorem.