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有兩種觀點來看什麼是 stochastic process：
第一種觀點關注的是單一 random variable 的資訊，像是 \(E[X_i], p_{X_i}(x)\) 等等；第二種觀點則關注整個序列，例如 \(P(X_i=1 \text{ for all } i)\)。
A Bernoulli process is a sequence of independent Bernoulli trials, \(X_i\), where \(X_i \sim \text{Ber}(p)\).
Independence and time homogeneity are the assumptions.
Let \(X_1, X_2, \cdots\) be a Bernoulli process, and let \(N\) be a random time. Then the process \(X_{N+1}, X_{N+2}, \cdots\) is also a Bernoulli process, which is independent of \(N\) and \(X_1, \cdots, X_N\).
不管在哪重新開始都一樣！
Let \(Y_k\) be the time of the \(k\)-th arrival, and let \(T_k\) be the \(k\)-th inter-arrival time, where \(T_k = Y_k - Y_{k-1}\). We can see that \(Y_k = T_1 + \cdots T_k\), and by the fresh start property, we know that every \(T_i\) are independent. Thus \(T_i \sim \text{Geo}(p)\).
畫個圖比較好想！
Moreover,
\[P(Y_k = t) = {t-1\choose k-1}p^k(1-p)^{t-k}.\]在 \(t-1\) 個 slots 中選 \(k-1\) 個；保留一次成功給 \(t\)。
見 Pascal r.v.。
Let \(X_1, X_2\cdots\) and \(Y_1, Y_2, \cdots\) be two Bernoulli processes, where \(X_i \sim \text{Ber}(p), Y_i \sim \text{Ber}(q)\). Then the merged process \(Z_1, Z_2, \cdots\) is also a Bernoulli process, where \(Z_i \sim \text{Ber}(p+q-pq)\).
If we split successes into two streams using independent flips of a coin with bias \(q\), we get two dependent Bernoulli processes.
Why dependent? Because a success in one process guarantees the failure in the other process.
\(\text{Ber}(p) \to \text{Ber}(pq), \text{Ber}(p(1-q))\).
假設 coin flip 和原 Bernoulli process 獨立。
畫個圖比較好想！