What can we conclude about the distribution of the sum of two random variables? If we know, for independent random variables $X$ and $Y$,
$P(X>x)\leq0.05$, and $P(Y>y)\leq0.05$, can we say anything about $P(X+Y>x+y)$? Can we be certain that it is less than $0.05$? Under what conditions could we say that?
 A: If $X$ and $Y$ are independent random variables such that 
we have $P\{X > x\} = a$ and $P\{Y > y\} = b$ where $x$, $y$, $a$, and $b$ are
numbers known to us, then
$$\begin{align*}
P\left(\{X > x\} \cup \{Y > y\}\right) &= P\{X > x\} + P\{Y > y\} 
- P\{X > x, Y > y\}\\
&= P\{X > x\} + P\{Y > y\} - P\{X > x\}P\{Y > y\}\\
&= a + b - ab.
\end{align*}$$
Now, the event $\{X+Y > x+y\}$ is a subset of the event 
$P\left(\{X > x\} \cup \{Y > y\}\right)$ and a superset of the event
$P\left(\{X > x\} \cap \{Y > y\}\right)$, and so we have that
$$ab \leq P\{X+Y > x+y\} \leq a + b - ab.$$
Both bounds are attainable.
Example: Take $X$ and $Y$ to be independent Bernoulli random
variables with parameter $\frac{1}{2}$. For $x=y=\frac{1}{4}$, 
we have
$$P\left\{X > \frac{1}{4}\right\}=P\left\{Y > \frac{1}{4}\right\}=\frac{1}{2};
~P\left\{X +Y > \frac{1}{2}\right\}= \frac{3}{4} = a+b-ab$$
while for $x=y=\frac{3}{4}$, 
we have
$$P\left\{X > \frac{3}{4}\right\}=P\left\{Y > \frac{3}{4}\right\}=\frac{1}{2};
~P\left\{X +Y > \frac{3}{2}\right\}= \frac{1}{4} = ab.$$
If all we know is that $P\{X > x\} \leq a$ and $P\{Y > y\} \leq b$
(that is, we only have upper bounds on the probabilities,
and the exact values of the probabilities might well be $0$),
then we cannot conclude that $ab \leq P\{X+Y > x+y\}$ since it might well
be that $P\{X+Y > x+y\} = 0$. But the upper bound
$$P\{X+Y > x+y\} \leq a + b -ab$$
still holds. Note that the complementary event $\{X+Y \leq x+y\}$ has
a subset $\{X\leq x, Y\leq y\}$ whose probability is
$$P\{X\leq x, Y\leq y\} = P\{X\leq x\}P\{Y \leq y\} \geq (1-a)(1-b)
= 1-a-b+ab$$ and so
$$P\{X+Y \leq x+y\}\geq 1-a-b+ab \Rightarrow P\{X+Y > x+y\}\leq a+b-ab.$$
