How prove $\sum_{j=1}^{n}x_{j}\prod_{i=j+1}^{n}(1-x_{i})=1-(1-x_{1})(1-x_{2})\cdots(1-x_{n})$ Let $x_{i}\in R$, show that
$$\sum_{j=1}^{n}x_{j}\prod_{i=j+1}^{n}(1-x_{i})=1-(1-x_{1})(1-x_{2})\cdots(1-x_{n})$$
I know this
$$1-(1-x_{1})(1-x_{2})=x_{1}+x_{2}+x_{1}x_{2}$$
But I can't prove my problem. 
Maybe there is a probability theory explanation?
 A: We can do this by induction:
$$\begin{align}
1 - (1 - x_1)(1 - x_2) \dots (1 - x_n) = & 1 - (1 - x_2) \dots (1 - x_n) + x_1 (1 - x_2) \dots (1 - x_n) \\
= & \sum_{j=2}^n x_j \prod_{i=j+1}^n (1 - x_i) + x_1 (1 - x_2) \dots (1 - x_n) \\
= & \sum_{j=1}^n x_j \prod_{i=j+1}^n (1 - x_i)
\end{align}$$
A probabilistic point of view: Assume that we have $n$ independent events $A_i$, each with probability $0 \le x_i \le 1$, $1 \le i \le n$. Now consider the event that at least one of $A_i$ happens. This is clearly $1 - (1 - x_1) \dots (1 - x_n)$. On the other hand we may express the union $\bigcup_{i} A_i$ as
$$A_n \cup (A_{n-1} \setminus A_n) \cup (A_{n-2} \setminus (A_n \cup A_{n-1}) \cup \dots \cup (A_1 \setminus (A_n \cup \dots \cup A_{2}))$$
This is a union of disjoint sets, which can also be written in the form
$$A_n \cup (A_{n-1} \cap A_n^c) \cup \dots \cup (A_1 \cap A_n^c \cap \dots \cap A_2^c)$$
and thus has the probability $\sum_{j=1}^n x_j \prod_{i=j+1}^n (1 - x_i)$.
A: For any index set $I$ we have 
$$\prod_{i\in I}(1-x_i)=\sum_{A\subseteq I} \prod_{i\in A}(-x_i),$$
where the sum runs over all subsets of $I$.
Applied to $I=\{1,2,\dots,n\}$ this gives 
$$\prod_{i=1}^n(1-x_i)=\sum_{A\subseteq \{1,2,\dots, n\}}
 \prod_{i\in A}(-x_i).$$
Now separate the  contributions from, first the empty set, and 
then those subsets of $\{1,2,\dots, n\}$ whose least 
element is $j$. Letting $A=\{j\}\cup B$, this gives
\begin{eqnarray*}\prod_{i=1}^n(1-x_i)&=&
\sum_{A\subseteq \{1,2,\dots, n\}} \prod_{i\in A}(-x_i)\\
&=&1+\sum_{j=1}^n (-x_j) \sum_{B\subseteq\{j+1,\dots, n\}} \prod_{i\in B}(-x_i)\\
&=&1-\sum_{j=1}^n x_j \prod_{i=j+1}^n(1-x_i).
\end{eqnarray*}
