The definition of Poisson binomial distribution is shown as https://en.wikipedia.org/wiki/Poisson_binomial_distribution, where $n$ independent trails with success probabilities $p_1,p_2,\ldots,p_n$. (The binomial distribution is a special case of the Poisson binomial distribution that $p_1=p_2=\cdots=p_n$.)
Let $X$ be a random variable following a Poisson binomial distribution, where $n$ independent trails with success probabilities $p_1,p_2,\ldots,p_i,\ldots,p_n$.
Suppose that we arbitrarily reduce a probability $p_i$ to $p'_i$ ($p_i > p'_i$), and have another random variable $Y$, which follows a Poisson binomial distribution with success probabilities $p_1,p_2,\ldots,p'_i,\ldots,p_n$.
Compared to the Poisson binomial distribution followed by $X$, the Poisson binomial distribution followed by $Y$ only has the same $p_1,\ldots,p_n$ except a lower $p'_i$. My goal is to show that $P(X\geq k) \geq P(Y \geq k)$, for any fixed $k \in \{1,\ldots,n\}$. In other words, I want to prove that $P(X\geq k)$ is a monotonic increasing function of $p_i$ for all $i = 1$ to $n$. The result looks simple, but I am struggling to prove that.
Note that $P(X\geq k) = \sum_{l=k}^n \sum_{A\in F_l} \prod_{i\in A} p_i \prod_{j\in A^c} (1-p_j) $ from Wikipedia. It is very hard to use an algebraic proof to show that $P(X\geq k) \geq P(Y \geq k)$. Can somebody give me a help?