# Independent Bernoulli random variables

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.

• 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. Oct 17, 2021 at 17:14
• @B.Liu sorry for my mistakes. I edited the post. Oct 18, 2021 at 4:47

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.

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