Question using Bonferroni’s inequality

combinatoricsprobability

The question is from Feller's probability volume 1

Question:

Independent events: $A_1, A_2, \dots \ , A_n$
$P\{A_k\} = p_k$

From Bonferroni's inequality, deduce that the probability of $k$ or more of the events $A_1,\dots,A_n$ occuring simultaneously is less than $\frac{(p_1 +\dots+p_n)^k}{k!}$

My approach: I have no approach. I solved the previous sub question afte struggling a lot but with this I have no idea where to begin. All I framed was ${\sum_{r=k}^{n}} {n \choose r}$. And this I guess something as to be done with it. I've spent over 2 hours on this but I can't make any progress.

This is not an assignment. I am teaching myself probability from Stat 110 on youtube and am doing this for my own practice. Any help is appreciated !

Best Answer

First, we introduce some useful notation: For any strictly increasing $k$-th tuple $J=(j_1,\ldots,j_k)$ with $1\leq j_1<\ldots<j_k\leq n$, let $$A_J :=A_{j_1}\cap\ldots\cap A_{j_k}$$ The event that that at least $k$ events out of $n$ events $\{A_1,\ldots,A_n\}$ happen is given by $$\bigcup_{J\in\mathcal{J}}A_J$$ where the index in the union ranges over all strictly increasing $k$-tuples in $\{1,\ldots, n\}^k$.

Notice that $$P[A_J]=P[A_{j_1}]\cdot\ldots\cdot P[ A_{j_k}]$$

Hence, $$\begin{align} P\Big[\bigcup_{J\in\mathcal{J}}A_J\Big]&\stackrel{1}{\leq}\sum_{J\in\mathcal{J}}P[A_J]\stackrel{2}=\sum_{1\leq j_1<\ldots<j_k\leq n}P[A_{j_1}]\cdot\ldots\cdot P[ A_{j_k}]\\ &\stackrel{3}{\leq}\frac{1}{k!}\big(P[A_1]+\ldots+P[A_n]\big)^k \end{align}$$

where:

  1. inequality (1) is Bonferroni's inequality,
  2. indentity (2) is due to independence,
  3. inequality (3) follows from the fact that each term of the for $P[A_J]$ with $J\in\mathcal{J}$ appears $k!$ times in the expansion of $\big(P[A_1]+\ldots+P[A_n]\big)^k$ ( I leave this combinatorial argument to the OP).
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