# Central limit theory [closed]

I know that when you keep adding Sn it will tend to approximate a normal curve as n gets bigger and bigger,but what happens if you change the expectancy ?

## closed as unclear what you're asking by kjetil b halvorsen, jbowman, AdamO, mdewey, FerdiSep 29 '18 at 22:46

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Roughly speaking, if we only change the expectation, then CLT will still hold, but each $$S_n$$ will be centered around $$\bar{\mu}_{n} = \frac{1}{n} \sum_{i=1}^{n} \mu_{i}$$. This is since CLT actually refers to the centered random variables. Note that if $$\bar{\mu}_{n} \to \infty$$, then anyway $$S_n \overset{P}{\to} \infty$$.

A more challenging extension is to change the variance. Then, the CLT doesn't always hold. A sufficient condition for it to hold is the Lyapunov condition. In essence, you need to add existence of a moment higher then second, and for it to converge slowly enough. There is some basic details in Wikipedia.

Edit: After the debate in the comment, I changed the note to convergence in probability. I wish to emphasize again that we deviate from the normal CLT assumptions only by changing the expectation. I provide here a proof for the claim:

et $$Y_n$$ be i.i.d with mean 0, variane $$\sigma^2$$ and assume third absolute moment exists ($$\mathbb{E}\left[\left|X\right|^3\right] < \infty$$). Let $$\mu_i$$ s.t. $$\frac{1}{n}\sum_{i=1}^{n} \mu_i \to \infty$$. Let $$X_i = Y_i + \mu_i$$. For every $$M>0$$, by Berry Essen theorem we have

$$\lim_{n\to\infty}P\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\ge M\right) =\lim_{n\to\infty}P\left(\frac{1}{\sqrt{n}\sigma}\sum_{i=1}^{n}Y_{i}\ge\frac{\sqrt{n}}{\sigma}\left(M-\frac{1}{n}\sum_{i=1}^{n}\mu_{i}\right)\right) \\ =\lim_{n\to\infty}P\left(-\frac{1}{\sqrt{n}\sigma}\sum_{i=1}^{n}Y_{i}\le\frac{\sqrt{n}}{\sigma}\left(\frac{1}{n}\sum_{i=1}^{n}\mu_{i}-M\right)\right) \\ \ge\lim_{n\to\infty}\Phi\left(\frac{\sqrt{n}}{\sigma}\left(\frac{1}{n}\sum_{i=1}^{n}\mu_{i}-M\right)\right)-\frac{C_{1}}{\sqrt{n}} \\ =\lim_{n\to\infty}1-\Phi^{C}\left(\frac{\sqrt{n}}{\sigma}\left(\frac{1}{n}\sum_{i=1}^{n}\mu_{i}-M\right)\right)-\frac{C_{1}}{\sqrt{n}} \\ \ge\lim_{n\to\infty}1-C_{2}e^{-\frac{n^{2}}{2\sigma}\left(\frac{1}{n}\sum_{i=1}^{n}\mu_{i}-M\right)^{2}}-\frac{C_{1}}{\sqrt{n}}=1$$

Where the last inequality hold since $$\frac{1}{n}\sum_{i=1}^{n}\mu_{i}-M$$ is positive eventually.

It seems possible to prove the almost surely convergence for sub-exponential random variables (using Borel-Cantelli lemma) but I'm not sure it holds in the general case.

• Re your note: this looks false. As a counterexample, let $Z_n$ be iid random variables of zero mean and unit variance. For $i=0, 1, 2, \ldots$ and $j=0, 1, \ldots, 2^i-1$ (representing $n=2^i+j$), define $X_n=i+2^{-i} Z_n.$ The $X_n$ are independent, the mean of their partial sums is asymptotically near $i$ and so diverges, yet the variance of their partial sums converges to $3.$ Thus if by "$S_n$" you mean the sample standard deviations of the partial sums, they will almost surely not diverge. – whuber Sep 30 '18 at 17:57
• @whuber $S_{n}$ marks the partial sums (not the sample sd of the partial sums, which i'd denote as lower case $s_n$). I meant $S_n = \sum_{k=1}^{n} X_{k}$ where $\left\lbrace X_{k} \right\rbrace_{k=1}^{\infty}$ are the series of r.v in question. I think that what OP meant by $S_{n}$ (judging from the wording) – tmrlvi Sep 30 '18 at 18:10
• In statements of the CLT, "$S_n$" often refers to the SDs used to standardize the partial sums. Your reference to "centered around" suggests this could be your meaning, too. This ambiguity will certainly lead to confusion. So that readers do not misunderstand you, you at least need to explain in your post what you mean by this symbol and why you think it's relevant to the question (which is so ambiguous it has been closed in the meantime). BTW, even in this changed sense your note is still incorrect! – whuber Sep 30 '18 at 18:20
• Here's a counterexample where $S_n$ are the partial sums. Let the $X_n$ be independent, taking on the value $3^n$ with probability $n/3^n$ and the value $0$ with probability $1-n/3^n.$ The partial sums of the means $\mu_n=n$ are $\bar\mu_n=(n+1)/2,$ which diverges, but there is a positive chance that all the $X_n$ are zero, implying $S_n=X_1+\cdots+X_n$ does not almost surely diverge. – whuber Sep 30 '18 at 18:25
• I think you are still confusing with the lower case $s_n$ that is often used as the normalized constant. Most references I encountered denoted $S_n$ just as I defined it. What references where you refering to? If you really want to get technical, CLT refers to normalizing by the true standard deviation (or mean standard deviation) not the sample mean. If I'm not mistaken, normalizing by the sample mean should converge to t-distribution. – tmrlvi Sep 30 '18 at 21:25