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Suppose that we have $n$ independent Bernoulli random variables, $x_1,\ldots,x_n$ such that each $x_i$ takes value 1 with success probability $p_i$ and value $0$ with failure probability $1-p_i$ ,i.e.,

\begin{align} P(x_1=1) & = p_1,\ P(x_1=0)= 1-p_1 \\ P(x_2=1) & = p_2,\ P(x_2=0) = 1-p_2 \\ & \vdots \\ P(x_n=1) & = p_n,\ P(x_n=0)=1-p_n \end{align}

So $x_1+...+x_n$ has Poisson binomial distribution since it is a sum of independent Bernoulli trials that are not necessarily identically distributed with the following pdf: $$ \mathsf{Pr}\left(K=k\right) = \sum_{A \in F_k} \prod_{i \in A} p_i \prod_{j \in A^c} (1-p_j) $$ where $F_k$ is the set of all subsets of k integers that can be selected from $\{1,\ldots,n\}$. and $A^c$ is the complement of A.

Suppose that I have calculated the following probability $$P(x_1+x_2+...+x_n<k)$$ Now I substitute one of $x_i$'s with another Bernoulli random variable (let say we substitute $x_1$ with $y_1$ with success probability $p'_1$) now I want to calculate $$P(y_1+x_2+...+x_n<k)$$ My question is : Can I use $P(x_1+x_2+...+x_n<k)$ to calculate $P(y_1+x_2+...+x_n<k)$ ?

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Write $z = x_2 + \ldots + x_n$. You have $$P(x_1 + z = k) = p_1 P(z = k-1) + (1-p_1) P(z=k)$$ so $$ \eqalign{P(z = 0) &= \frac{P(x_1 + z = 0)}{1-p_1} \cr P(z = 1) &= \frac{P(x_1 + z = 1) - p_1 P(z = 0)}{1 - p_1} = \frac{P(x_1 + z = 1)}{1-p_1} - \frac{p_1 P(x_1 + z = 0)}{(1 - p_1)^2}\cr P(z = 2) &= \frac{P(x_1 + z = 2)}{1-p_1} - \frac{p_1 P(x_1+z=1)}{(1-p_1)^2} + \frac{p_1^2 P(x_1 + z = 0)}{(1-p_1)^3}\cr P(z = k) &= \sum_{j=0}^k \frac{(-p_1)^j P(x_1 + z = k-j)}{(1-p_1)^{j+1}}}$$ and then $$P(y+z = k) = p'_1 P(z=k-1) + (1-p'_1) P(z=k)$$

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