# Central Limit Theorem with (linear) weights

I have a question about the CLT. Suppose we have the independently and identically distributed random variables $X_i$ with mean $\mu$ and variance $\sigma^2$. Then, by the Central Limit Theorem

$$\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n X_i-\mu\right) \rightarrow N\left(0, \sigma^2\right)\Longleftrightarrow \sum_{i=1}^n X_i \rightarrow N(n\mu, n\sigma^2),$$ as $n\rightarrow \infty$. But what if we have $\sum_{i=1}^nc_iX_i$, where $c_i$ are constants? Is it simply to say that

$$\sum_{i=1}^nc_iX_i \rightarrow N\left(E\left[\sum_{i=1}^nc_iX_i\right], \;V\left[\sum_{i=1}^nc_iX_i\right]\right),$$

that is,

$$\sum_{i=1}^nc_iX_i \rightarrow N\left(\mu\sum_{i=1}^nc_i,\; \sigma^2\sum_{i=1}^nc_i^2\right),$$

or am I missing something here?

• What is the meaning of a convergence in distribution to a normal distribution with variable mean and variance? The question applies to the convergence to $N(n\mu,n\sigma^2)$ at the beginning of the question as well as to the convergence you are trying to establish later on. – Did Apr 14 '13 at 11:35
• I'm not sure I understand your question. A sum (and a sample mean) converges to $n$ times variable mean and variance, no? Or do you mean that it has to be expressed in terms of convergence to a standard normal? – hejseb Apr 14 '13 at 12:12
• In probability as in other branches of mathematics, the statement that $x_n\to y_n$ has no meaning. The convergence of a sequence $(x_n)$ is expressed as $x_n\to y$, for some $y$ which does not depend on $n$. The two statements my previous comment addresses both fail this requirement. – Did Apr 14 '13 at 12:16
• Ah, yes, of course. That makes sense and is something I didn't think about. Thank you for your help, and excuse my mathematical recklessness! – hejseb Apr 14 '13 at 12:24

You might be after central limit theorems à la Lindeberg or Lyapunov, see here. These yield the convergence you are interested in, but stated more correctly, for example, some conditions ensuring that the sequence of general term $$\frac{\sum\limits_{i=1}^nc_iX_i-\mu\sum\limits_{i=1}^nc_i}{\sigma\cdot\sqrt{\sum\limits_{i=1}^nc_i^2}}$$ converges in distribution to a standard normal distribution.