- I flip $N$ coins.
- The $n^{th}$ coin flip has probability $p_n$ to be head.
- If the $n^{th}$ coin flip is head, I add $c_n$ to the total sum S.
How to calculate/approximate the expectation of $S$ quickly? Is it possible to simplify the following?
The number of coin flip is about 10.
Let $\vec{c} = (c_1, c_2, …, c_N)$
Let $\vec{a} = (a_1, a_2, …, a_N)$ be the result of the coin flips.
$a_n$ is 1 if the $n^{th}$ coin flip is head and is 0 otherwise.
The expectation is:
$E\lbrace S \rbrace = \sum_{a_{1}, … a_{n}} \vec{c} \cdot \vec{a} \prod_{n = 1}^{N} a_n p_n + (1 – a_n)(1 – p_n) $
But that involves enumerating all cases of $\vec{a}$, which is not quite practical for my case because I need to calculate this for many times.
The $p_n$ are different for different coins.
Best Answer
By linearity of expectation, the expected value of the sum is
$$ \mathsf E[S]=\mathsf E\left[\vec c\cdot\vec a\right]=\vec c\cdot\mathsf E\left[\vec a\right]=\vec c\cdot\vec p\;. $$