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For two events $A$ and $B$, then how to prove that

$$\Pr(A+B\geq 0) \leq \Pr(A\geq 0) + \Pr(B\geq 0)$$


I have a proof as follows, but I think it is not quite intuitive which needs to write it down to figure it out.

\begin{align} \Pr(A+B\geq 0) &= 1-\Pr(A+B < 0)\\\\ &\leq 1-\Pr(A<0 \cap B < 0)\\\\ & = \Pr(A\geq 0 \cup B \geq 0)\\\\ & \leq \Pr(A\geq 0) + \Pr(B\geq 0) \end{align} The second inequality by the observation that $Pr(A+B < 0) \geq \Pr(A<0 \cap B < 0)$, and the last step by union bound.

But this can not provide an intuition for me, is there simper proof or way to figure it out?

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  • 3
    $\begingroup$ If A and B are events how can they be compared to a number like 0? Also the + symbol is strange. Are you using it to mean the intersection of events? $\endgroup$ – Michael Chernick Apr 24 '17 at 13:22
  • $\begingroup$ You are right, i think it has to be reformulated. $\endgroup$ – Peng Zhao Apr 25 '17 at 0:14
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You need to reformulate this in terms of random variables $X$ and $Y$.

The reasoning is that if $X+Y\geq 0$, then either $X\geq 0$ or $Y\geq 0$; it can't be that both $X$ and $Y$ are negative. Hence, the event $$ \{X+Y\geq 0\} $$ is a subset of $$ \{X\geq 0\}\cup\{Y\geq 0\}. $$ Since $\Pr(\;\cdot\;)$ is a monotonic set function, we have $$ \Pr\{X+Y\geq 0\} \leq \Pr\left(\{X\geq 0\}\cup\{Y\geq 0\}\right)\leq\Pr\{X\geq 0\}+\Pr\{Y\geq 0\}, $$ in which the last inequality follows from the subadditivity of $\Pr(\;\cdot\;)$.

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As Zen says, you need to talk of random variables and not events.

$\{X+Y \geq 0\}$ is the event that the random point $(X,Y)$ lies on or above the diagonal line $x+y=0$ through the origin, which divides the plane into to two half-planes:the upper half-plane $\{(x,y)\colon x+y \geq 0\}$ and the lower half-plane $\{(x,y)\colon x+y < 0\}$. Draw a sketch of the plane with coordinate axes $x$ and $y$ and mark on it the line $x+y=0$ to see why the half-planes are called so. In fact, mark the line $x=y$ on your sketch too and note that you have divided the plane into eight octants (numbered $1$ through $8$) and that the region of interest (the upper half-plane) is the disjoint union of the first, second, third, and eighth octants. Let $p_i$, $1 \leq i \leq 8$, denote the probability that $(X,Y)$ lies in the $i$-th octant. Then, we have that $$P\{X+Y \geq 0\} = p_1+p_2+p_3+p_8\tag{1}$$ Now, the half-plane $\{(x,y)\colon x \geq 0\}$ is the disjoint union of the first, second, seventh and eighth octants, while the half-plane $\{(x,y)\colon y \geq 0\}$ is the disjoint union of the first through the fourth octants. Hence, \begin{align} P\{X \geq 0\} &= p_1+p_2+p_7+p_8,\tag{2}\\ P\{Y \geq 0\} &= p_1+p_2+p_3+p_4.\tag{3} \end{align} Consequently, \begin{align} P\{X + Y\geq 0\} &= p_1+p_2+p_3+p_8,\\ &\leq (p_1+p_2+p_3)+p_4 + (p_8) + p_1+p_2+p_7\\ &= (p_1+p_2+p_3+p_4) + (p_1+p_2+p_7+p_8)\\ &= P\{Y \geq 0\} + P\{X \geq 0\}\\ \text{that is,} ~~P\{X + Y\geq 0\} &\leq P\{X \geq 0\} + P\{Y \geq 0\}. \tag{4} \end{align} Furthermore, equality can hold in $(4)$ if and only if $p_1=p_2=p_4=p_7 = 0$.

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