Proving that as N increases, it is more likely that x - y > z, with x \in X, y \in Y, z \in Z

Question

In trying to solve a bigger applied problem, I found myself facing the following.

Let $X$, $Y$ and $Z$ be three independent random variables, each coming from an unknown distribution, and each with $1...n$ realizations $x \in X, y \in Y, z \in Z$ . I know that:

• they all have the same number $n$ of sampled realizations.
• they all have the same support, i.e. $x,y,z \in [a,b]$, where $a,b \in \mathbb{R^{\geq0}}$ and $b>a$.

So, let's say we draw $n$ samples from each variable, so we get $n$ realizations from each of the three variables (where each draw is indexed by $i$ such that $i \in 1...n$). How can I show that as $n$ increases, it becomes more likely that $x_i + y_i > z_i$ for at least one draw $i$? If extra assumptions are needed, which would be the minimal assumptions required?

Answer 1

Given a sequence of independent events $A_i$ each of them has probability $p > 0$, the probability of having at least one event $A_i$ happening for $i \le n$, is $1-(1-p)^n$ which is always increasing and converges to $1$. So you only have to prove that $P(X+Y > Z) > 0$.

Your assumptions about $X, Y$ and $Z$ are a bit unclear, but if there is a couple of values $x, y$ for which:

$$P(X > x) > 0,\\ P(Y>y) > 0,\\P(Z < x+y) > 0$$

then, there is a probability greater than 0 that $X+Y >Z$, and what you want to prove is proven already.