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A sequence of Bernoulli trials satisfies the following assumptions:

  1. Each trial has two possible outcomes, in the language of reliability called success and failure.

  2. The trials are independent. Intuitively, the outcome of one trial has no influence over the outcome of another trial.

  3. On each trial, the probability of success is $p$ and the probability of failure is $1−p$ where $p \in [0,1]$ is the success parameter of the process.

http://www.math.uah.edu/stat/bernoulli/Introduction.html

How is independence defined in this case? Is it pairwise independence or mutual independence? The probability of particular outcome ($k$ successes after $n$ trials) is given by $$p^k (1-p)^{n-k}$$

It implies we have to assume mutual independence (which is a stronger assumption than pairwise independence).

Let $S_i$, $i\in \mathbb{N}$, $1\le i \le n$ denote the event that we got success in $i$th trial, and $F_i$ - failure.

Therefore we require that (it's just one of many other conditions required by mutual independence) $$P(A_1 \cap A_2 \cap A_3)=P(A_1)P(A_2)P(A_3)$$

Obviously to talk about events, we need an appropriate sample space. A single outcome is a sequence of length $n$ of $S$ and $F$s. The event $S_i$ corresponds to all outcomes that have $S$ in the $i$th place.

Is there anything that should be added here? Do I understand it correctly?

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    $\begingroup$ When left unspecified, independence usually refers to mutual independence. A sample space convenient for an unspecified number of trials is $\{0,1\}^\mathbb N$ the set of (infinite) sequences with values in $\{0,1\}$, where the $n$th entry of the sequence being $0$ or $1$ means the $n$th try is a failure or a success, respectively. $\endgroup$ – Did Jan 13 '16 at 11:06
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Your understanding is correct. We require mutual independence here. If we only have pairwise independence, we can let $A3 = A1 \mathbin{\oplus} A2$, where $\mathbin{\oplus}$ denotes xor. Then you can verify pairwise independence because \begin{align*} P(A_1 = a, A_3 = b) &= P(A_1 = a, A_2 = a \mathbin{\oplus} b) \\ &= P(A_1 = a)P(A_2 = a \mathbin{\oplus} b) = P(A_1 = a) P(A_3 = b) \end{align*} But this is not a Bernoulli process

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