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If $X_n$ converges to zero in distribution, do $\frac{1}{\sqrt{n}} X_n$ and $\frac{1}{n} X_n$ converge to zero in distribution? More generally, if $\lim a_n = 0$, do we have $a_n X_n \overset{d}\to 0$?

I set out to prove this but I got little confused on what is the cdf of zero. I am attempted to use Slutsky Theorem, but $a_n$ is not really a sequence of random variables?...

2 Answers2

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This is a direct conclusion by applying Slutsky's theorem.

Ѕᴀᴀᴅ
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    You're not addressing OP's concerns. He specifically mentioned that "$a_n$ is not really a sequence of random variables". You should add a few lines explaining why it is the case. – Gabriel Romon Dec 31 '17 at 13:39
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$X_n$ converges to zero in distribution iff it converges to zero in probability.

If you want a proof of that then have a look here or here.

So we have $\lim_{n\to\infty}\mathsf P(|X_n|\geq\epsilon)=0$ for every $\epsilon>0$.

If $a_n\to0$ then eventually $\{|a_nX|\geq\epsilon\}\subseteq\{|X_n|\geq\epsilon\}$ so that $\mathsf P(|a_nX|\geq\epsilon)\leq\mathsf P(|X_n|\geq\epsilon)$.

So indeed we get: $\lim_{n\to\infty}\mathsf P(|a_nX_n|\geq\epsilon)=0$ for every $\epsilon>0$ and conclude that $a_nX_n$ converges to zero in distribution.

drhab
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