Central limit theorem replacing radical n with n

Question

Does $n(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2/ n^2),$ remembering that $n^{1/2}(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2)?$

Answer 1

There are 3 issues

1) notation $\overset{\mathbb P}{\to}$ versus $\overset{\mathbb d}{\to}$

^{This you have already edited/corrected in your question.}

The notation $\overset{\mathbb P}{\to}$ means convergence in probability.

With the central limit theorem we use, the notation $\overset{\mathbb d}{\to}$, which means convergence in distribution.

2) The variance $n\sigma^2$ versus $\sigma^2/n^2$

Possibly you meant to write $$n(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,n\sigma^2)$$

Instead of

$$n(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2/n^2)$$

The asymptotic behaviour of the variance is $$\text{Var}\left( n(y_n-\mu)\right) \sim n \sigma^2$$

3) Convergence versus asymptotic equivalence

Convergence relates to a limit, like the distribution function approaching some other function.

$X_n \overset{\mathbb P}{\to} \mathcal f(x) $

means that for all values of $x$ we have the limiting behaviour:

$$\forall x: \left\{ \lim_{n \to \infty} P(X_n<x) = f(x) \right\}$$

Or in your example

$$\forall y: \left\{\lim_{n \to \infty} P\left(\frac{Y_n-\mu}{\sigma n ^{1/2}} \leq y \right) = \Phi(y) \right\}$$

A limit/convergence, using the notation $\to$, has always a fixed constant value or function on the right side. When the right side is changing, then we speak about asymptotic behaviour. This is denoted with a tilde, $\sim$, see for instance the Wikipedia article about this topic definition of asymptotic equility relationship (not to be confused with the tilde notation in statistics).

It wouldn't be bad to have the right side change as well (although you would need find a different, less confusing, notation for it) and we could say that it means:

$$\tag{1}\lim_{n \to \infty} \frac{P\left(Y_n \leq y \right)}{\Phi\left( \frac{y-\mu}{n^{1/2}\sigma} \right)}= 1 $$

But this is a different statement than

$$\tag{2} \lim_{n \to \infty} \frac{P\left(\frac{Y_n-\mu}{\sigma n ^{1/2}} \leq y \right)}{ \Phi(y)} = 1$$

You may have that equation 1 does not hold for every value of $y$ while equation 2 does hold for every value of $y$.

Example:

In the image below you see the distribution of the mean of 100 Bernoulli distributed variables. It does not approach a normal distribution in the sense that the limit of the ratio goes to 1. Close to zero this ratio remains zero (and this doesn't get better when we increase $n$).

In your case,

If

• you would define the equivalence relationship

  $$f_n(x) \approx g_n(x)$$

  as denoting asymptotic equivalence for the two distribution functions and more specifically as

  $$\forall x: \left\{\lim_{n \to \infty} \frac{F_n(x)}{G_n(x)} = 1 \right\}$$

  where $F$ and $G$ relate to the cumulative distribution functions generated by the probability density functions $f$ and $g$.

Then

• You can state that $$n(y_n-\mu) = \sum_{i=1}^n (y_i-\mu) \approx \mathcal N(0,n\sigma^2)$$

It is not common, but this expression for asymptotic normality is not entirely absent. For example in Barndorff-Nielsen, Ole E., and David Roxbee Cox. Asymptotic techniques for use in statistics.

The sequence $\lbrace V_n \rbrace$ is asymptotically normal if there exist sequences of constants $\lbrace a_n \rbrace, \lbrace b_n\rbrace$ such that $(V_n - a_n)/b_n $ converges in distribution to the standard normal distribution. The constants $a_n, b_n$ are called respectively the asymptotic mean and standard deviation of $V_n$.

We sometimes write $V_n \sim aN(a_n, b_n^2)$. Asymptotic normality is concerned with...

Although this does not define $\sim$ very clearly and may be more like an abuse of notation and $V_n \sim aN(a_n, b_n^2)$ means $(V_n-a_n)/b_n \sim N(0, 1)$ and not the asymptotic equivalence relationship that I described before.

Answer 2

Notice a few things.

1. If we compute the variance of the LHS, $\text{Var}(n(y_n-y)) = n^2\text{Var}(y_n)$. However, $y_n = \frac{1}{n}(x_1+...+x_n)$ where the $(x_i)_{1\leq i\leq n}$ are i.i.d. (I am assuming based on your notation). Therefore, $\text{Var}(y_n) = \tfrac{1}{n^2}(n\cdot \sigma^2) = \tfrac{\sigma^2}{n}$. We conclude from here that the variance of the LHS is equal to $n\cdot \sigma^2$. As $n\to \infty$ this quantity blows up to infinity and there is no limit.
2. As $n\to \infty$ the quantity $\frac{\sigma^2}{n^2}\to 0$, therefore you are suggesting that the distribution of $n(y_n-y)$ converges in distribution to $N(0,0)$ which looks very suspicious.

Answer 3

No, write $$n(y_n-\mu)=n^{1/2}n^{1/2}(y_n-\mu)$$ From $n^{1/2}(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2)=O_p(1)$ (cf. e.g. root-n consistent estimator, but root-n doesn't converge?) we find $$n(y_n-\mu)=O_p(n^{1/2})$$ we obtain something which diverges whereas, as Nicolas has pointed out, $N(0,\sigma^2/ n^2)$ tends to a degenerate r.v. at zero, so basically the constant zero.