Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $X,Y$ be independent random variables, uniform on $(0,1)$.

a) $P(X+Y>1.5)$.

b) $P(X>Y \mid X>1/2)$.

c) $P(\tan^{-1}(Y/X)<t)$ for all $0<t<\pi/2$.

e) $E(\tan^{-1}(Y/X))$.

I could use the definitions I learned about joint distribution to solve part (a). But I am not sure how to approach the second one when there is conditional probability involves. And for the last two parts, I don't even know how to solve them when there is function like $\tan$ involves.

share|cite|improve this question
up vote 2 down vote accepted

HINT: a), c) and e) are direct uses of probability and expectation definitions for joint distributions.

b) You need to also use the definition of conditional probability $$ \mathbb{P}\left( X > Y ; Y > \frac{1}{2} \right) = \frac{\mathbb{P}\left( X > Y \land Y > \frac{1}{2} \right)}{ \mathbb{P}\left(Y > \frac{1}{2} \right) } = \frac{\mathbb{P}\left( X > Y > \frac{1}{2} \right)}{ \mathbb{P}\left(Y > \frac{1}{2} \right) } $$

share|cite|improve this answer

Try thinking about $\int_0^1\int_0^1 h(x,y)\;\;dx\;\;dy$ where $h(x,y)$ is a function that is true when $x>y | x>1/2$.

share|cite|improve this answer

These images should help with: you want the red area divided by the sum of the red and pink areas. For the third, the form of the answer may change depending on $t \lt \pi / 4$ and $t \gt \pi / 4$.

enter image description here

The expectation should be easy to spot using symmetry.

share|cite|improve this answer
that is a really nice way to visualize the problem. May I ask what kind of software you used to generate these graphs? – geraldgreen Nov 5 '11 at 20:19
@John: MS Paint – Henry Nov 5 '11 at 20:20

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.