# Confusion about joint distribution between two independent random variables

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.

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## 3 Answers

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 $$\mathbb{P}\left( X > Y ; Y > \frac{1}{2} \right) = \frac{\mathbb{P}\left( X > Y \land Y > \frac{1}{2} \right)}{ \mathbb{P}\left(Y > \frac{1}{2} \right) } = \frac{\mathbb{P}\left( X > Y > \frac{1}{2} \right)}{ \mathbb{P}\left(Y > \frac{1}{2} \right) }$$

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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 $x>y | x>1/2$.

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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 $t \gt \pi / 4$.

The expectation should be easy to spot using symmetry.

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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