<s>[INST] Probability of a sum of probabilistic random variables? Suppose we have $\mathbb P(A > x) \leq m$ and $\mathbb P(B > x) \leq m$.  What is $\mathbb P(A + B > y)$?  I have been looking for a related axiom and not had any luck.
 [/INST] Without any additional assumptions (such as independence), we can say the following:
If $x > \frac{1}{2}y$, we can't bound $\mathbb P(A + B > y)$. In fact, for any $p$, we can find $A$ and $B$ such that $\mathbb P(A + B > y) = p$ regardless of the value of $m$. Otherwise, when $x ≤ \frac{1}{2}y$, we can infer that $\mathbb P(A + B > y) ≤ m$.
To prove this, first consider the case when $x > \frac{1}{2}y$. Let $p ∈ [0, 1]$. Set $ε = \frac{1}{2}(x - \frac{1}{2}y)$. Give $(A, B)$ the following joint distribution:


*

*With probability $p$, yield $(\frac{1}{2}y + ε,\; \frac{1}{2}y + ε)$,  

*With probability $1 - p$, yield $(\frac{1}{2}(y - 1),\; \frac{1}{2}(y - 1))$.


Then $A + B$ has the following distribution:


*

*With probability $p$, yield $\frac{1}{2}y + ε + \frac{1}{2}y + ε = y + 2ε$.

*With probability $1 - p$, yield $\frac{1}{2}(y - 1) + \frac{1}{2}(y - 1) = y - 1$.


Thus, $\mathbb P(A + B > y) = p$. To check that $\mathbb P(A > x) ≤ m$ (and, similarly, that $\mathbb P(B > x) ≤ m$), notice that $A$ is surely at most $\frac{1}{2}y + ε$, and
$$\begin{align}\tfrac{1}{2}y + ε
&= \tfrac{1}{2}y + \tfrac{1}{2}(x - \tfrac{1}{2}y) \\
&= \tfrac{1}{2}y + \tfrac{1}{2}x - \tfrac{1}{4}y \\
&= \tfrac{1}{2}x + \tfrac{1}{4}y \\
&< \tfrac{1}{2}x + \tfrac{1}{2}x \\
&= x,\end{align}$$
so $P(A > x) = 0 ≤ m$.
Now consider the case when $x ≤ \frac{1}{2}y$. Whenever $A + B > y$, $A + B > 2x$, so either $A > x$ or $B > x$. Thus, either $\mathbb P(A + B > y) ≤ \mathbb P(A > x)$ or $\mathbb P(A + B > y) ≤ \mathbb P(B > x)$, so $\mathbb P(A + B < y) ≤ m$.
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