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

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?

• do X, Y and Z have the same distribution? If they don't, your claim is not generally true. Nov 11 '19 at 22:54
• @carlo Ops, I see. I edited the question to ask for showing that as $n$ increases, at it increases the probability that for at least one draw from the variables, it will be true.
– ZXiu
Nov 11 '19 at 23:03
• ok. pay attention to sign, in any case. for b negative that can't be true Nov 11 '19 at 23:07
• Your language is still ambiguous. You say "three i.i.d. random variables of unknown distributions." Do you mean that you have three distributions for $X,Y,Z$ and you're drawing iid samples from each? Nov 11 '19 at 23:19
• @AlexR. I mean exactly that each of those variables follows one distribution, possibly different from each other. The 3 variables are independent from each other. I always draw samples from the 3, but again, the sampled realizations are independent form each other.
– ZXiu
Nov 11 '19 at 23:52

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.