# Show that $\max\{\textbf{P}((A\cup B)^{c}),\textbf{P}(A\cap B),\textbf{P}(A\triangle B)\}\geq\frac{4}{9}$

Let $$A$$ and $$B$$ be independent events. Show that \begin{align*} \max\{\textbf{P}((A\cup B)^{c}),\textbf{P}(A\cap B),\textbf{P}(A\triangle B)\}\geq\frac{4}{9} \end{align*}

MY ATTEMPT

Since $$\textbf{P}(A\cap B) = \textbf{P}(A)\textbf{P}(B)$$, we have

\begin{align*} \textbf{P}((A\cup B)^{c}) & = 1 - \textbf{P}(A\cup B) = 1 - \textbf{P}(A) - \textbf{P}(B) + \textbf{P}(A\cap B)\\ & = 1 - \textbf{P}(A) - \textbf{P}(B) + \textbf{P}(A)\textbf{P}(B)\\ & = (1 - \textbf{P}(A)) - \textbf{P}(B)(1 - \textbf{P}(A))\\ & = (1 - \textbf{P}(A))(1-\textbf{P}(B)) = \textbf{P}(A^{c})\textbf{P}(B^{c}) \end{align*}

Analagously, we get \begin{align*} \textbf{P}(A\triangle B) & = \textbf{P}(A) + \textbf{P}(B) - 2\textbf{P}(A\cap B) = \textbf{P}(A) + \textbf{P}(B) - 2\textbf{P}(A)\textbf{P}(B)\\\\ & = \textbf{P}(A)(1 - \textbf{P}(B)) + \textbf{P}(B)(1-\textbf{P}(A)) = \textbf{P}(A)\textbf{P}(B^{c}) + \textbf{P}(A^{c})\textbf{P}(B) \end{align*}

However, I do not know how to proceed from here. Am I on the right track? Could someone complete the proof? Any help is appreciated. Thanks in advance.

• Why not write $p=P(A)$ and $q=P(B)$? It would make manipulating your formulae a lot easier. – Angina Seng Jan 12 '19 at 4:42
• Try to fix, for example, $P(A\cap B) < \frac{4}{9}$, and write the sum of the probabilities of the other two events you are considering. This sum has to be greater than some quantity. Can you than deduce, from that, your thesis? – dfnu Jan 12 '19 at 11:54

As in the comment by Lord Shark The Unknown you can first set $$P(A) = p$$ and $$P(B) = q.$$ Then the thesis becomes the following.
Show that the inequalities $$\begin{eqnarray} pq &<& \frac{4}{9}\\ (1-p)(1-q) &<& \frac{4}{9}\\ p(1-q) + q(1-p) &<& \frac{4}{9} \end{eqnarray}$$ cannot hold simultaneously.
Performing the following change of variables $$$$\begin{cases} p = \frac{1}{2}(Q+P+1)\\ q = \frac{1}{2}(Q-P+1) \end{cases}$$$$ leads to the corresponding inequalities in terms of $$P$$ and $$Q$$, i.e. $$\begin{eqnarray} (Q+1)^2 -P^2 &<& \frac{16}{9}\tag{1}\label{uno}\\ (Q-1)^2 -P^2 &<& \frac{16}{9}\tag{2}\label{due}\\ Q^2-P^2>\frac{1}{9}\tag{3}\label{tre} \end{eqnarray}$$ which are areas delimited by hyperbolae (in red, green, and black respectively in the Figure). The red dashed area corresponds to the set of points $$(P,Q)$$ satisfying conditions \eqref{uno} and \eqref{due}. The black dashed areas mark te set of points that satisfy condition \eqref{tre}. Therefore the three inequalities cannot be all true. And the thesis follows.
• @user1337, I noticed that expressing the conditions with respect to $(p,q)$ lead to curves that had a nicer representation once rotated by $\frac{\pi}{4}$ and with origin shifted by $\left(\frac{1}{2}, \frac{1}{2}\right)$, resulting in the substitution I proposed you. Maybe there are nicer way to approach the problem, of course. – dfnu Jan 13 '19 at 19:18
• @user1337, as a matter of fact, a rotation of $\frac{\pi}{4}$ should be enough to "see" the result. That is, ignoring the scaling factor, $P=p-q$ and $Q=p+q$. – dfnu Jan 13 '19 at 19:45