# Does this Bonferroni styled inequality also hold for characteristic functions?

This is the popular Bonferroni inequality. Does it also hold for characteristic functions of random variables, as in when $P(A_i)$ is replaced by the characteristic function $\chi(A_i)$ and so forth?

Given random variables $A_1,A_2...A_n$ (not events like in Theorem 1.2.3), is this satisfied? $$\sum_{i=1}^n||\chi_{A_i}(t)|| - \sum_{i <j}^n||\chi_{A_i|A_j}(t).\chi_{A_j}(t)|| \le ||\chi_{\sum_{i=1}^n A_i}(t)||$$

• What is the definition of your characteristic function? Apparently (?), it is something different than en.wikipedia.org/wiki/… , which everyone in probability knows and loves, because that characteristic function is associated with a random variable, not with an event $A_i$ as you seem to have. Or perhaps you are rather confused. In any event, what is your "proposed" inequality? – Mark L. Stone Dec 1 '17 at 1:34
• Given random variables $A_1,A_2...A_n$ (not events like in 1.2.3), and $$\sum_{i=1}^n||\chi(A_i)|| - \sum_{i <j}^n||\chi(A_i|A_j).\chi(A_j)|| \le ||\chi(\sum_{i=1}^n A_i)||$$ is satisfied? So..the second term has conditional characteristic fn. and $A_1..A_n$ are not necessarily independent. If you want to first check for the case where they are independent as a special case, you are welcome to as well. – hearse Dec 1 '17 at 2:11
• I still have no idea what $\chi(A_i)|$ or $||\chi(A_i)||$ or $\sum_{i=1}^n A_i$ mean. I don't know whether anyone else knows what they mean. – Mark L. Stone Dec 1 '17 at 2:14
• The norm of a characteristic function of r.v's is always $\le$ 1. Check in say. en.wikipedia.org/wiki/… bullet no. 4 in properies. It is well-defined. – hearse Dec 1 '17 at 2:15
• Updated........ – hearse Dec 6 '17 at 22:29

Example: $n=2$, take random variables $X=1_A$ and $Y=1_B$ where $A$ and $B$ are independent events. Then $\chi_X(t)=e^{it}P(A)+P(A^c)$, so we can clearly choose $t$ such that $x:=|\chi_X (t)|<1$. Define also $y:=|\chi_Y(t)|$. In this situation, the stated inequality is equivalent to $x+y-xy\leq xy$ i.e. $x+y\leq 2xy$. Combining this with the AM-GM inequality leads to $\sqrt{xy}\leq xy$, which is clearly false when $xy<1$.