Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|improve this question
up vote 8 down vote accepted

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

share|improve this answer
I think you put a $\cup$ in place of a $\cap$ when writing the subset/superset bit. – guy Dec 21 '12 at 6:49
@guy Thanks for the careful reading. I corrected the typo and also improved the bound. – Dilip Sarwate Dec 21 '12 at 12:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.