central limit theoremlaw-of-large-numbers
I had a question in my mind , if a i.i.d distribution function follows central limit theorem , does that mean it will follow Strong law of large numbers also ??
Since in both cases sample means tends to population mean
Please explain .
Here i mean Lindberg levy's Clt .
The OP says
The central limit theorem states that the mean of i.i.d. variables, as N goes to infinity, becomes normally distributed.
I will take this to mean that it is the OP's belief that for i.i.d. random variables $X_i$ with mean $\mu$ and standard deviation $\sigma$, the cumulative distribution function $F_{Z_n}(a)$ of $$Z_n = \frac{1}{n} \sum_{i=1}^n X_i$$ converges to the cumulative distribution function of $\mathcal N(\mu,\sigma)$, a normal random variable with mean $\mu$ and standard deviation $\sigma$. Or, the OP believes minor re-arrangements of this formula, e.g. the distribution of $Z_n - \mu$ converges to the distribution of $\mathcal N(0,\sigma)$, or the distribution of $(Z_n - \mu)/\sigma$ converges to the distribution of $\mathcal N(0,1)$, the standard normal random variable. Note as an example that these statements imply that $$P\{|Z_n - \mu| > \sigma\} = 1 - F_{Z_n}(\mu + \sigma) + F_{Z_n}((\mu + \sigma)^-) \to 1-\Phi(1)+\Phi(-1) \approx 0.32$$ as $n \to \infty$.
The OP goes on to say
This raises two questions:
- Can we deduce from this the law of large numbers? If the law of large numbers says that the mean of a sample of a random variable's values equals the true mean μ as N goes to infinity, then it seems even stronger to say that (as the central limit says) that the value becomes N(μ,σ) where σ is the standard deviation.
The weak law of large numbers says that for i.i.d. random variables $X_i$ with finite mean $\mu$, given any $\epsilon > 0$, $$P\{|Z_n - \mu| > \epsilon\} \to 0 ~~ \text{as}~ n \to \infty.$$ Note that it is not necessary to assume that the standard deviation is finite.
So, to answer the OP's question,
The central limit theorem as stated by the OP does not imply the weak law of large numbers. As $n \to \infty$, the OP's version of the central limit theorem says that $P\{|Z_n-\mu| > \sigma\} \to 0.317\cdots$ while the weak law says that $P\{|Z_n-\mu| > \sigma\} \to 0$
From a correct statement of the central limit theorem, one can at best deduce only a restricted form of the weak law of large numbers applying to random variables with finite mean and standard deviation. But the weak law of large numbers also holds for random variables such as Pareto random variables with finite means but infinite standard deviation.
I do not understand why saying that the sample mean converges to a normal random variable with nonzero standard deviation is a stronger statement than saying that the sample mean converges to the population mean, which is a constant (or a random variable with zero standard deviation if you like).
This figure shows the distributions of the means of $n=1$ (blue), $10$ (red), and $100$ (gold) independent and identically distributed (iid) normal distributions (of unit variance and mean $\mu$):
As $n$ increases, the distribution of the mean becomes more "focused" on $\mu$. (The sense of "focusing" is easily quantified: given any fixed open interval $(a,b)$ surrounding $\mu$, the amount of the distribution within $[a,b]$ increases with $n$ and has a limiting value of $1$.)
However, when we standardize these distributions, we rescale each of them to have a mean of $0$ and a unit variance: they are all the same then. This is how we see that although the PDFs of the means themselves are spiking upwards and focusing around $\mu$, nevertheless every one of these distributions is still has a Normal shape, even though they differ individually.
The Central Limit Theorem says that when you start with any distribution--not just a normal distribution--that has a finite variance, and play the same game with means of $n$ iid values as $n$ increases, you see the same thing: the mean distributions focus around the original mean (the Weak Law of Large Numbers), but the standardized mean distributions converge to a standard Normal distribution (the Central Limit Theorem).
Best Answer
In short: No.
A bit longer: Convergence in distribution does not directly imply, by any way, convergence almost-surely.
Much longer:
Given a random vector $X$, whose components are independent and identically distributed and its first two moments are finite - then the CLT says that asymptotically, the sample mean $\bar{X}_n$ converges to $\mu$ with rate $\sqrt{n}$ and asymptotic distribution $N(0,\sigma^2)$. That's convergence in distribution. It means that the CLT provides us with information regarding the rate in which the sample mean converges weakly to the population mean as sample size increases.
The Weak LLN says that the sample mean of a vector with a finite first moment, converges in probability to the population mean. That is, $\lim_{n\rightarrow\infty}P(\left| \bar{X}_n-\mu \right|>\epsilon)=0$. This means that no matter how small is the nonzero margin $\epsilon$ that we take, a sufficiently large sample would bring the difference between the sample mean and the population mean inside this margin. As you can see here, it is possible (under several conditions) that the CLT would imply the WLLN.
The Strong LLN says that the sample mean converges almost surely to the population mean. That is, $P\left(\lim_{n\rightarrow\infty}\bar{X}_n=\mu \right)=1$. This means that for a sufficiently large sample size, the probability of $\bar{X}_n$ not converging to $\mu$ is 0. That is a substantially stronger form of convergence, and cannot be directly implied from the CLT.