I'm not really understanding this Taylor expansion approximation. It states that it uses the fact that $2|\epsilon_i| < 1$ and that it is a first order approximation. The $P_i$ is a probability, so it is also $< 1$ Can someone give me some direction here?
$$ \frac12 \prod_{i=1}^m \left(1+2\epsilon_i\right)^{P_i} \approx\frac12 + \sum_{i=1}^m P_i \epsilon_i $$
On
This can be derived from the Binomial Theorem which states that $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} = a^n + nab^{n-1} + \ldots, $$ and in case $b$ is very small, the other terms with higher orders of powers of $b$ vanish, so you have the approximation $$ (1 + b)^n \approx 1 + bn $$ for small $b$. Note the original theorem assumes $n$ to be a positive integer, but can be extended.
Using the first term of the Taylor series of $(1+x)^{\alpha}$, that is $$(1+x)^{\alpha} = 1 + \alpha x + O(x^2)$$ which works in the region $\vert x \vert < 1$. We can say that $$(1+2 \epsilon_i)^{P_i} = 1 + 2 P_i \epsilon_i + O(\epsilon_i^2)$$ only when $\vert 2 \epsilon_i \vert < 1$. Do the product over $N$ terms, $$ \prod_{i=1}^m (1+2 \epsilon_i)^{P_i} \simeq \big( 1 + 2 P_1 \epsilon_1 + O(\epsilon_1^2)\big ) \big( 1 + 2 P_2 \epsilon_2 + O(\epsilon_2^2)\big) \ldots \big( 1 + 2 P_m \epsilon_m + O(\epsilon_m^2) \big)$$ now neglecting all term higher than order $1$ and dividing by $2$ on both sides, we are done, $$ \frac{1}{2} \prod_{i=1}^m (1+2 \epsilon_i)^{P_i} \simeq \frac{1}{2} + \sum_{i=1}^m P_i \epsilon_i$$