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If $X_n $ converges in probability to $X$, does $\dfrac{X_n}{n}$ converge in probability to $\dfrac{X}{n}$?

I wish to apply to this to see if $\dfrac{s}{\sqrt{n}}$ converges in probability to $\dfrac{\sigma}{\sqrt{n}}$ given that $s$ converges in probability to $\sigma$.

Let $\rightarrow$ imply convergence by probability.

First Method I used:

I know that $X_n \rightarrow X$ and $Y_n\rightarrow Y$, then $X_nY_n \rightarrow XY$. Thus, since $Y_n=\dfrac{1}{n} \rightarrow0$, does this imply that $X_n \dfrac{1}{n} \rightarrow X(0)=0$? Using this method I see that$\dfrac{X_n}{n}$ does not converge in probability to $\dfrac{X}{n}$.

Second Method I used

Looking at the definition of convergence by probability, I obtain for $n\ge1$:

$P(|X_n/n-X/n|>\varepsilon)=P(\dfrac{|X_n-X|}{n}>\varepsilon)\le P(|X_n-X|>\varepsilon)$

Taking the limit, the RHS of the inequality will equal to $0$ by assumption and hence the LHS will also equal to $0$, implying that $\dfrac{X_n}{n}$ converges in probability to $\dfrac{X}{n}$.

Both methods seemingly contradict each other.

shahril effendi
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2 Answers2

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The statement "$X_n/n$ converges in probability to $X/n$" does not have any precise meaning, because the claimed limit, $X/n$, depends on $n$.

It is true that

  • $X_n/n\stackrel{p}{\to}0$ and $X/n\stackrel{p}{\to}0$ so the two sequences have the same limit (your first approach)
  • $X_n/n-X/n\stackrel{p}{\to}0$ (your second approach)

These aren't especially interesting results because the same would hold for $X_n/n$ having the same limit as $2X/n$ or $-17X/n$ or whatever.

Given your motivation of relating $s/\sqrt{n}$ to $\sigma/\sqrt{n}$, you might be better off looking at whether $X_n/X\stackrel{p}{\to} 1$ implies $$\frac{X_n/\sqrt{n}}{X/\sqrt{n}}\stackrel{p}{\to} 1$$ It does (trivially; just cancel the $\sqrt{n}$s), and this sort of result lets you exchange $s/\sqrt{n}$ and $\sigma/\sqrt{n}$ when they are used as denominators.

Thomas Lumley
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If you are looking at $$\lim\limits_{n \to \infty} \frac{X_n}{n}$$ then $n$ is a bound variable and should not appear in the limit. So on that basis your first approach is correct, and your second is wrong.

If you are interested in the Central Limit Theorem and you see something like $$\lim\limits_{n \to \infty} \frac{\bar X_n - \mu}{\frac{s}{\sqrt{n}}}$$ converging in distribution to $\mathcal N(0,1)$, you actually have a numerator with $\lim\limits_{n \to \infty} {\bar X_n - \mu} =0$ and $\lim\limits_{n \to \infty} \frac{s}{\sqrt{n}}=0$, which is why you are interested in the limit of the quotient rather than the $\frac00$ quotient of the limits.

It still allows you to estimate the standard error of the sample mean with $\frac{s}{\sqrt{n}}$, but this tends to reduce towards $0$ as $n$ increases, as you might naturally hope with an increasing sample size.

Henry
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