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My professor left us to solve this problem:

Let $\xi_1, \xi_2,...,\xi_n$ be independent Bernoulli random variables defined on a probability space $(\Omega,P(\Omega),\mathbb{P})$ such that:

$\mathbb{P}(\xi_i=0) = 1 -\lambda_i\Delta$, $\mathbb{P}(\xi_i=1) = \lambda_i\Delta$, where $\lambda_1, \lambda_2,...,\lambda_n$ are positive, and $0<\Delta<\frac{1}{max\{\lambda_1,\lambda_2,...\lambda_n\}}$.

Prove that $\mathbb{P}(\xi_1+\xi_2+...+\xi_n = 1) \geq \Delta\sum_{i=1}^{n}\lambda_i - \Delta^2(\sum_{i=1}^n\lambda_i)^2$.

I can prove that $\mathbb{P}(\xi_1+\xi_2+...+\xi_n = 1) \leq \Delta\sum_{i=1}^{n}\lambda_i$ since $\mathbb{P}(\xi_1+\xi_2+...+\xi_n = 1) = \sum_{i=1}^n\lambda_i\Delta(\Pi_{j=1\\j\neq i}^n1-\lambda_j)^2 \leq \Delta\sum_{i=1}^n\lambda_i$ but I don't know how to prove another inequality.

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    $\begingroup$ Welcome to CV, Igalala! Please edit your question to include the self-study tag (which you can do by clicking "edit" at lower left). Self-study question are treated differently on CV, and your question will receive better attention when you do this. $\endgroup$
    – Alexis
    Oct 17, 2021 at 17:14
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    $\begingroup$ @B.Liu sorry for my mistakes. I edited the post. $\endgroup$
    – Igalala
    Oct 18, 2021 at 4:47

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You can rewrite

$$\Delta\sum_{i=1}^{n}\lambda_i - \Delta^2(\sum_{i=1}^n\lambda_i)^2 = \sum_{i=1}^{n} \left[\Delta \lambda_i \cdot \left( 1 - \sum_{i=1}^n \Delta\lambda_i\right) \right] $$

And compare with the exact expression

$$\sum_{i=1}^{n} \left[\Delta \lambda_i \cdot \prod_{j\neq i}\left( 1 - \Delta\lambda_j\right) \right] $$

This product $ \prod_{j\neq i}\left( 1 - \Delta\lambda_j\right)$ expressing the probability that the other $\xi_j = 0$ when $\xi_i =1$ is simplified with a sum that underestimates this value.

This is similar to the difference between the Bonferroni correction and the Šidák correction. You can proof this with Boole's inequality.

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  • $\begingroup$ Oh... I was thinking about writing this instead of trying to apply Boole's inequality: $\Pi_{j\neq i}(1-\Delta\lambda_j) = 1-\Delta\sum_{j\neq i} \lambda_j + \lambda^2\sum_{j_1,j_2\neq1}\lambda_{j_1}\lambda_{j_2} - ... \geq 1 - \sum_{i=1}^n\delta\lambda_i$. $\endgroup$
    – Igalala
    Oct 18, 2021 at 10:23

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