1. ## Two random variables...

P1. Lets say we have 2 random variables X~Uniformly(0,1) and Y~Uniformly(0,2)
Determine the probability P(X<Y).
I suspect the answer is 1/4, but I do now know how to formally get to this result.

P2. the same as above but this time X~Exp(1) and Y~Exp(2)

P1. Lets say we have 2 random variables X~Uniformly(0,1) and Y~Uniformly(0,2)
Determine the probability P(X<Y).
I suspect the answer is 1/4, but I do now know how to formally get to this result.

P2. the same as above but this time X~Exp(1) and Y~Exp(2)

$\displaystyle \displaystyle P(X<Y)=\int_{x=0}^1\int_{y=x}^2 f(x,y)\ \;dydx$

CB

3. That makes sense. Thanks a lot!

4. and for the second one I'd use

$\displaystyle \int_0^{\infty}\int_0^y f(x,y)dxdy$

5. So... who should f(x,y) be? In the first case I saw there was no need for f(x,y)
From what I understood the second should be:

$\displaystyle \int_0^{\infty}\int_{e^y}^{\infty} 2*e^{-2*x}dxdy$

Is this right?

6. you need to review densities
and you need f(x,y) in both cases
in the first problem

f(y)=1/2 and f(x)=1 since these are uniforms
By indep the joint density is the product.

in the second question $\displaystyle f_X(x)=e^{-x}$
and the density of Y can eithe be $\displaystyle f_Y(y)=2e^{-2y}$ or $\displaystyle (1/2)e^{-y/2}$

depending how you define the exponential distribution.

7. Ok, I know who $\displaystyle f_X(x)=e^{-x}$ and $\displaystyle f_Y(x)=2e^{-2x}$ are, but how do I get from them to the $\displaystyle f(x,y)$ mentioned above?

Sorry, I probably look like a big noob, but I really need to know these to pass an exam. Thx for your time

Ok, I know who $\displaystyle f_X(x)=e^{-x}$ and $\displaystyle f_Y(x)=2e^{-2x}$ are, but how do I get from them to the $\displaystyle f(x,y)$ mentioned above?
$\displaystyle f_{X,Y}(x,y)=f_X(x)f_Y(y)$