Bonferroni's Inequality

## Bonferroni's Inequality

For $n$ events $A_1,A_2,\cdots, A_n$

` $$ \begin{equation}\label{bof} P\big(\cap_{i=1}^n A_i\big)\geq \sum_{i=1}^n P(A_i) -(n-1). \end{equation} $$ `

#### Proof

For $n =2$,

` $$ \begin{equation*} P(A_1\cup A_2)= P(A_1)+P(A_2) -P(A_1\cap A_2) \end{equation*} $$ `

But $P(A_1\cup A_2)\leq 1$. Using this in above equation, we have

` $$ \begin{eqnarray*} & &P(A_1\cup A_2)= P(A_1)+P(A_2) -P(A_1\cap A_2)\leq 1 \\ \implies & & P(A_1\cap A_2) \geq P(A_1) +P(A_2) -1. \end{eqnarray*} $$ `

Hence from above inequality, it is clear that the inequality \eqref{bof} is true for $n=2$.

Suppose the inequality \eqref{bof} is true for $n=r$.

` $$ \begin{equation}\label{bof2} P\big(\cap_{i=1}^r A_i\big)\geq \sum_{i=1}^r P(A_i) -(r-1). \end{equation} $$ `

Then

` $$ \begin{eqnarray*} P\big(\cap_{i=1}^{r+1} A_i\big)& = & P\big(\cap_{i=1}^r A_i \cap A_{r+1}\big)\\ & \geq & P\big(\cap_{i=1}^r A_i\big) + P( A_{r+1}) -1\\ &\geq & \sum_{i=1}^r P(A_i) -(r-1)+ P(A_{r+1}) -1\\ & & \text{ (Using inequality \eqref{bof2})}\\ & \geq &\sum_{i=1}^{r+1} P(A_i) -(r+1-1). \end{eqnarray*} $$ `

Hence inequality \eqref{bof} is true for $n=r+1$. So by mathematical induction, the inequality \eqref{bof} is true for all $n$.

## Bonferroni's Inequality Example

If $P(A) = 0.9$ and $P(B) = 0.8$, show that $P(A \cap B)\geq 0.7$.

#### Solution

As $A\cup B \subseteq S$, we have $P(A\cup B) \leq P(S) = 1$.

But

` $$ \begin{eqnarray*} P(A\cup B) &\leq& 1\\ P(A) + P(B)- P(A\cap B)& \leq & 1\\ P(A\cap B) &\geq & P(A) +P(B) -1 \end{eqnarray*} $$ `

Thus

` $$ \begin{eqnarray*} P(A\cap B) &\geq & P(A) +P(B) -1\\ P(A\cap B) & \geq & 0.9 +0.8 -1\\ &\geq &0.7. \end{eqnarray*} $$ `

## Conclusion

In this tutorial, you learned about Bonferroni's Inequality and how to prove it.

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