# Do these two equalities hold?

$$\mathbb P (x+y > N) = \mathbb P \left(x > \frac N2 \right) + \mathbb P \left(y > \frac N2 \right) \tag{1}$$

$$\mathbb P (\mid x+y-z \mid> N) = \mathbb P \left(\mid x-z \mid > \frac N2 \right) + \mathbb P \left(\mid y \mid > \frac N2 \right) \tag{2}$$

They seem to me unreasonable since for neither of them the two events on the left hand side of the equality are mutually exclusive. If there is one theorem similar to the above expressions, could anyone kindly show me? Many thanks in advance!

• This is indeed wrong. Consider for instance two variables with always take positive values and $N=0$. This cannot possibly be right then, since you would get $1=1+1$. – Wojowu Apr 19 at 13:22
• An interesting question would be to ask when the equality does hold. – Rodrigo de Azevedo Apr 19 at 13:33

Let $$x$$ and $$y$$ be uniformly distributed on $$[N,N]$$. (Yes, one point). Then your equation says $$1 = 1 + 1$$, which seems unlikely.
Now let $$x$$ be distributed uniformly on $$[N/2+1, N/2+1]$$ and $$y$$ be distributed uniformly on $$[0,0]$$. Then your equation is $$0 = 0+1$$.
Now suppose that $$x$$ and $$y$$ are both uniformly distributed on $$[0,N]$$. The two probabilities on the right are $$1/2$$, as is the probability on the left. (Draw a $$[0,N] \times [0,N]$$ square. The probability on the left hand side is the fraction of the square above the line $$x+y = N$$, which is half the area. The probabilities on the right hand side are the right half and upper half of the area of the square, respectively.) Then your equation says $$1/2 = 1$$.
In fact, the diagram above suggests the truth : the set of outcomes with $$x + y > N$$ is a subset of the union $$U = \{(x,y): x > N/2\} \cup \{(x,y): y > N/2\}$$. Therefore \begin{align*} &P(x+y > N) \\ &\leq P(x > N/2) + P(y > N/2) - P(x > N/2 \text{ and } y > N/2) \\ &\leq P(x > N/2) + P(y > N/2) \text{,} \end{align*} where the middle uses inclusion-exclusion to only count the contributions from $$U$$ once.