Show that $\mathsf P\left[\bigcap_iA_i\right]\ge\sum_i\mathsf P(A_i)-(n-1)$

Question

Let $A_1, A_2, \dots$ be events. Show that for all $n\ge1$ $$\mathsf P\left[\bigcap_iA_i\right]\ge\sum_i\mathsf P(A_i)-(n-1)$$

I am able to prove this for $n=2$ but I need to prove it for all $n$. When $n=2$ there's events A and B with $$\mathsf P(A\cap B)\ge\mathsf P(A)+\mathsf P(B)-(2-1)$$

which may be seen from the following relations: $$\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)$$ $$\mathsf P(A\cap B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cup B)$$ $$P(A\cup B)≤1$$

To prove for all $n$ I think I need to use inclusion–exclusion, but where does the $n-1$ come from?

No problem. Have you tried induction? Assuming it is true for ${A_1,\cdots, A_{n-1}}$ let $A=\bigcap_{i=1}^{n-1}A_i$ and let $B=A_n$ and use your two state argument. — lulu, Oct 24 '16 at 01:18

score 1 · Accepted Answer · answered Oct 24 '16 at 01:21

After rearranging a lot of terms, the inequality you want becomes $$ \sum_i(1-P(A_i))\geq1-P(\bigcap_iA_i). $$ Now use that, for any event $X$, $1-P(X)=PX^c)$ (where the superscript $c$ means complement) and that $(\bigcap_iA_i)^c =\bigcup_i({A_i}^c)$.

Show that $\mathsf P\left[\bigcap_iA_i\right]\ge\sum_i\mathsf P(A_i)-(n-1)$

1 Answers1

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