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i need to create an example of identically distributed, but dependent Bernoulli random variables where $ x_1,x_2....x_n$ i.e $x\in{0,1}$ such that, $$P\big(|\mu-\frac{1}{n}\sum_{i=1}^{n}x_i|\geq \frac{1}{2}\big)=1$$

where $\mu=E[x_i]$,

The example should show that independence is crucial for convergence of mean to the expected values. i.e $\mu=E[x_i]$

im a non mathematics student struglling a bit with understanding the concept.i was wondering how $\mu$ equals $E[x_i]$ and its relation to independence. a short explanation will really help.

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  • $\begingroup$ So, did any of the two answers help you figure it out? $\endgroup$ – Clement C. Dec 3 '19 at 19:07
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Take $(x_1,...,x_n)$ such that $x_1$ is a Bernoulli r.v. with parameter $1/2$, and $x_1=x_2=\dots=x_n$ a.s. Then $\mu=1/2$, but $$ \frac{1}{n}\sum_{i=1}^n x_i = x_1\in\{0,1\} $$ so $ \left\lvert \mu - \frac{1}{n}\sum_{i=1}^n x_i \right\rvert = 1/2 $ a.s.

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  • $\begingroup$ im a non mathematics student struglling a bit with understanding the concept.i was wondering how $\mu$ equals $E[x_i]$ and its relation to independence. a short explanation will really help. $\endgroup$ – Balash Dec 9 '19 at 1:57
  • $\begingroup$ @Balash $\mu= \mathbb{E}[x_i]$ by definition. Since each $x_i$ is Bernoulli with parameter $1/2$, $\mathbb{E}[x_i] = 0\cdot 1/2 + 1\cdot 1/2 = 1/2$. Independence (or lack therefore) doesn't change the expectation of each fixed $x_i$. What it changes is stuff like the covariance of $x_i$ and $x_j$, for instance. $\endgroup$ – Clement C. Dec 9 '19 at 2:24
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Let $x_1$ be chosen at random (with $\mu=0.5$) and $x_{k+1}=1-x_k$ for all other $k$. These are obviously dependent and satisfy the condition on $P$.

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