What can we conclude about the distribution of the sum of two random variables?

Question

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?

Answer 1

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 $$P\{X+Y \leq x+y\}\geq 1-a-b+ab \Rightarrow P\{X+Y > x+y\}\leq a+b-ab.$$