Probability: $N$ dependent Bernoulli trials?

Question

I understand the proof, but I don't see why Bernoulli trials are mutually independent [...] is not necessary.

Answer 1

There is a more elegant way to prove this. Making use of linearity of expectation we find:$$\mathbb E(B_1+\cdots+B_n)=\mathbb EB_1+\cdots+\mathbb EB_n=np$$ since $\mathbb EB_i=P(B_i=1)=p$ for every $i\in\{1,\dots,n\}$.

Linearity of expectation also works if the $B_i$ are not independent. So this route is not only (much) more elegant but also shows that independence is not necessary.

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Advice: If $X$ has binomial distribution with parameters $n$ and $p$ then always think of it as a sum $X=B_1+\cdots+B_n$ where the $B_i$ are iid and have Bernoulli distribution with parameter $n$. In lots of situations that makes things more easy. This is one these situations.

Answer 2

$E[ \sum_1^n X_i ] = \sum_1^n E[ x_i] = \sum_1^n E[ x_1] = n E[x_1] = np$