I've heard that there is a soft version or interpretation of the CLT that says —in summary— that it can be applied also to a sequence of independent real-valued random variables that share the same distribution shape but not the mean (for instance, a sequence of $n$ normal variables $X_i \sim N(\mu_i, \sigma)$ where $\mu_i$ can be different for each $i$).

Is that right? Where can I find it (textbook or article)?

And, a last one: In this case (where $X_i$ are independently and equally distributed except for their means), what would the distribution of $Y=\frac{1}{n}\sum_{i=1}^n{X_i}$ approximately be, according to CLT?


I want to stress that $X_i\sim N(\mu_i,\sigma)$ was just an example.

What I have is $X_i \sim F_i$, where $F_i$ stands for the CDF of $X_i$, and where all $F_i$'s are equal to the same distribution (with a known constant variance $\sigma^2$) except for the fact that they have different averages $\mu_i$.


Just to make it clearer, what I would like to know is whether or not it is correct to state this:

$Y=\frac{1}{n}\sum_{i=1}^n{X_i}$ is approximately distributed as $N(\mu_Y,\sigma_Y)$, where

  • $\mu_Y$ can be estimated as $\bar{x} = \frac{x_1+ \cdots + x_n}{n}$ and
  • $\sigma_Y$ can be estimated as $\frac{s}{\sqrt{n}}$, $s$ being the sample (quasi)standard deviation,

using a random sample $(x_1, \dots, x_n)$ of $(X_1, \dots, X_n)$.

  • 1
    $\begingroup$ What do you know about $\mu_i$? $\endgroup$
    – Emre
    Jun 11, 2015 at 7:34
  • $\begingroup$ @Emre, that is a good question. All what we know about $\mu_i$ is that it is not constant. We can assume that there is a kind of underlying central $\mu$ which could be estimated as the average $\bar{x}=\frac{x_1+\dotsc+x_n}{n}$, where $x_i$ is an observation from $X_i$. $\endgroup$
    – Vicent
    Jun 11, 2015 at 7:48
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    $\begingroup$ You don't need a CLT for independent normal variables; $X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1 ^2+ \sigma_2^2)$ and so on. $\endgroup$
    – KOE
    Jun 11, 2015 at 7:51
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    $\begingroup$ Google "Lindeberg Feller central limit theorem" $\endgroup$ Jun 11, 2015 at 8:04
  • 1
    $\begingroup$ See these Wikipedia links 1 2 $\endgroup$
    – Glen_b
    Jun 11, 2015 at 10:34

1 Answer 1


A precise reference for the Lindeberg-Feller CLT is Theorem 4.12 in Foundations of Modern Probability by Olav Kallenberg. I paraphrase it below:

Let $(\xi_{n,j}; j = 1,\ldots,n)_{n=1}^\infty$ be a triangular array of row-wise independent random variables with mean $0$ and $\sum_{j=1}^n \mathbb{E}[\xi_{n,j}^2] \to 1$ as $n\to\infty$.

Suppose that for any $\epsilon > 0$ we have $\sum_{j=1}^n \mathbb{E}[\xi_{n,j}^2; |\xi_{n,j}|>\epsilon] \to 0$. (*)

Then $\sum_{j=1}^n \xi_{n,j} \overset{d}{\to} N(0,1)$.

Suppose the $X_i$ are arbitrary independent random variables. Let $\mu_j = \mathbb{E}[X_j]$ and $\sigma_j^2 = Var[X_j]$.

Take $\xi_{n,j} = \frac{X_j - \mu_j}{\sqrt{\sum_{i = 1}^n \sigma_i^2}}$. Then the theorem says $$ \frac{1}{\sqrt{\sum_{i = 1}^n \sigma_i^2}} \sum_{j = 1}^n (X_j - \mu_j) \overset{d}{\to} N(0,1), $$ and thus $$ \frac{1}{\sqrt{n}} \sum_{j=1}^n (X_j - \mu_j) \overset{d}{\to} N(0,\sigma^2) $$ if the limit $\sigma^2 = \lim_{n\to\infty}\sum_{i=1}^n \sigma_i^2 / n$ exists. To put in another way, the sample mean $Y_n = \frac{1}{n} \sum_{i=1}^n X_i$ satisfies $$ \sqrt{n}(Y_n - \frac{1}{n}\sum_{i=1}^n \mu_i) \overset{d}{\to} N(0,\sigma^2), $$ i.e. $Y_n$ has approximate distribution $N(\frac{1}{n}\sum_{i=1}^n \mu_i, \sigma^2/n)$. However, be warned that the theorem does not say exactly what the rate of convergence is!

I leave it as an exercise for you to see what the Lindeberg condition (*) is.

  • $\begingroup$ Therefore, just to be sure, does this mean that the distribution of $Y=\frac{1}{n}\sum_{i=1}^n{X_i}$ asymptotically tends to a normal distribution with mean $\mu=\frac{1}{n}\sum_{i=1}^n{E[X_i]}$ and standard deviation $\sigma= \frac{\sqrt{\sum_{i=1}^n{Var[X_i]}}}{n} $? See also my last edit above, please. $\endgroup$
    – Vicent
    Jun 11, 2015 at 13:45
  • $\begingroup$ Your $\mu$ and $\sigma$ here are not limiting values... do you mean $\mu = \lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^n \mathbb{E}[X_i]$? (assuming the limit exists, and similar for $\sigma$) $\endgroup$ Jun 11, 2015 at 15:58
  • $\begingroup$ So, $\mu_Y = \lim_{n \rightarrow +\infty}{\frac{1}{n}\sum_{i=1}^n{E[X_i]}}$ and $\sigma_Y = \lim_{n \rightarrow +\infty}{\frac{1}{n}\sqrt{\sum_{i=1}^n{Var[X_i]}}}$ would be right, in case those limits exist. Isn't that? $\endgroup$
    – Vicent
    Jun 11, 2015 at 16:04
  • $\begingroup$ And, about my last edition in the original question, could $\mu_Y$ and $\sigma_Y$ be approximated by $\bar{x}$ and $\frac{s}{\sqrt{n}}$ from a sample $(x_1,\dots,x_n)$ of $(X_1,\dots,X_n)$ ?? $\endgroup$
    – Vicent
    Jun 11, 2015 at 16:08
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    $\begingroup$ More or less- see expanded answer. $\endgroup$ Jun 11, 2015 at 22:01

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