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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?

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  • $\begingroup$ do X, Y and Z have the same distribution? If they don't, your claim is not generally true. $\endgroup$
    – carlo
    Nov 11 '19 at 22:54
  • $\begingroup$ @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. $\endgroup$
    – ZXiu
    Nov 11 '19 at 23:03
  • $\begingroup$ ok. pay attention to sign, in any case. for b negative that can't be true $\endgroup$
    – carlo
    Nov 11 '19 at 23:07
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    $\begingroup$ 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? $\endgroup$
    – Alex R.
    Nov 11 '19 at 23:19
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    $\begingroup$ @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. $\endgroup$
    – ZXiu
    Nov 11 '19 at 23:52
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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.

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    $\begingroup$ What do you mean by those three probabilities being greater than 1? $\endgroup$
    – ZXiu
    Nov 12 '19 at 0:38
  • $\begingroup$ sorry, I meant greater than 0. just corrected it $\endgroup$
    – carlo
    Nov 12 '19 at 3:17

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