What rule is used to expand the following product (LHS) into the expression (RHS)?

Question

$\prod_{j=1}^{m} (1-e^{- r_j \cdot x}) = 1 - \Sigma_{i=1}^{m} e^{-r_i\cdot x } + \Sigma_{i=1}^{m} e^{-r_i\cdot x}\cdot \Sigma_{j>i}^{m}e^{-r_j\cdot x} - \Sigma_{i=1}^{m}e^{-r_i\cdot x}\cdot \Sigma_{j>i}^{m}e^{-r_j\cdot x} \cdot \Sigma_{k>j}^{m}e^{-r_k\cdot x} + \cdots + (-1)^m e^{-\Sigma_{i=1}^{m}r_i\cdot x}$

Answer 1

The Principle of Inclusion-Exclusion.

We can solve this by making an analog with probability theory.

Since multiplication is analogous to $\cap$, and addition/subtraction is analogous to inclusion/exclusion, we might as well be finding:

$$|\bigcap_{j=1}^m (X - A_j)|,$$

where $X$ is the universal set.

Now, to see the right hand side, we can tentatively include $|X|$, but we have overcounted, so we can take away the sum of $|A_j|$, getting

$$|X| - \sum_{j=1}^m |A_j|.$$

However, we have oversubtracted those elements that are in two or more of the $A_j$’s, so we add them back. Iteratively, we get that the right hand side should be:

$$|X| - \sum_{j=1}^m |A_j| + \bigg(\sum_{j=1}^m |A_j|\bigg)\bigg(\sum_{k=1}^m |A_k| \bigg)- \ldots + (-1)^m \underbrace{\sum \cdots \sum |A_{\cdots}|}_{\prod_{i=1}^m |A_i|}.$$

Now, converting back to your equation means that the right hand side now makes sense.

Answer 2

Let $x>0$, $0\leq A_i=e^{-r_i \cdot x} \leq 1$.

As $A$ is probability, now we can write $1-A = \bar A$. Products are analogous to intersections therefore we can write

\begin{equation} \tag{1} \prod_{i=1}^{m} P(1 - e^{-r_i \cdot x}) =\prod_{i=1}^{m} P(\bar A_i) = \bigcap_{i=1}^{m} P(\bar A_i) = 1 - \bigcup_{i=1}^{m} P(A_i) \end{equation}

Now we can use inclusion-exclusion principle

\begin{equation} \bigcup_{i=1}^{m} P(A) = \Sigma_{i=1}^{m} P(A_i) - \Sigma_{1\leq i < j \leq m} P(A_i \cap A_j) + \cdots + (-1)^m (A_i \cap A_{i+1} \cap \cdots \cap A_m) \end{equation}

We convert intersections back to products

\begin{equation} \bigcup_{i=1}^{m} P(A) = \Sigma_{i=1}^{m} P(A_i) - \Sigma_{i= 1}^{m}P(A_i)\cdot\Sigma_{j>i}^ {m} P(A_j) + \cdots + (-1)^m P(A_i)\cdot P(A_{i+1}) \cdots P(A_m) \end{equation}

Insert values of $Ai$s, i.e., $A_i=e^{-r_i \cdot x}$

\begin{equation} \bigcup_{i=1}^{m} P(A) = \Sigma_{i=1}^{m} e^{-r_i \cdot x} - \Sigma_{i= 1}^{m}e^{-r_i \cdot x} \cdot\Sigma_{j>i}^ {m} e^{-r_j \cdot x} + \cdots + (-1)^m e^{-r_i \cdot x}\cdot e^{-r_{i+1} \cdot x} \cdots e^{-r_m \cdot x}\end{equation}

Further simplification yield

\begin{equation} \bigcup_{i=1}^{m} P(A) = \Sigma_{i=1}^{m} e^{-r_i \cdot x} - \Sigma_{i= 1}^{m}e^{-r_i \cdot x} \cdot \Sigma_{j>i}^{m} e^{-r_j \cdot x} + \cdots + (-1)^m e^{-\Sigma_{i=1}^{m}{r_i \cdot x}} \end{equation}

Put this equation back into (1) we get \begin{equation} \prod_{i=1}^{m}(1-e^{r_i\cdot x}) = 1 - \Sigma_{i=1}^{m} e^{-r_i \cdot x} - \Sigma_{i= 1}^{m}e^{-r_i \cdot x} \cdot \Sigma_{j>i}^{m} e^{-r_j \cdot x} + \cdots + (-1)^m e^{-\Sigma_{i=1}^{m}{r_i \cdot x}} \end{equation}