The information given in the task:
*X and Y are independent random variables.
$\ X$~$R(0,1)$
$\ Y$~$R(0,1)$*
I am then told to calculate:
$P(X>1/4)$, expected values and so on. The last problem is the one I am struggling with:
Calculate $P(X+Y>1|Y>1/2)$, (The solution is 3/4)
Somehow I'm unable to solve this. My idea is:
Given that both X and Y are uniformly distributed I know that their PDF's are:
$\ f_Y(y)=\frac{1}{b-a}, x \in [a,b]$
$\ f_X(x)=\frac{1}{b-a}, x \in [a,b]$
Since;
$\ X$~$R(0,1)$
$\ Y$~$R(0,1)$
we get:
$\ f_Y(y)=\frac{1}{1-0}, x \in [0,1]$
$\ f_X(x)=\frac{1}{1-0}, x \in [0,1]$
Since this problem is dealing with conditional I thought I would use the formula
$\ f_{X|Y}(x|y)=\frac{f_{XY}(x,y)}{f_Y(y)}$
Since $f_Y(y)$ is already calculated I need $f_{XY}(x,y)$
Since X and Y are independent the definition of this independence gives me:
$f_Y(y)·f_X(x)=f_{XY}(x,y)$
So:
$f_{XY}(x,y)=\frac{1}{1-0}·\frac{1}{1-0}=1$
This means:
$\ f_{X|Y}(x|y)=\frac{f_{XY}(x,y)}{f_Y(y)}=\frac{1}{1}=1$
My idea is then to use the formula
$P(A|B)=\frac{P(A \cap B)}{P(B)}$
$P(B)=\int_{1/2}^{1}\frac{1}{1-0}dy=\frac{1}{2}$
But how do I proceed from here when neither of the pdf's have any variables in them? Given that the integrals that I need to solve has to depend on X or Y I keep getting the wrong solutions when trying to solve it. In other words, im stuck.
$P(X+Y>1,Y>1/2)=P(X>1-Y,Y>1/2)=\int_{\frac{1}{2}}^{1}\int_{1-y}^{1}f_{XY}(x,y)dx dy=\int_{\frac{1}{2}}^{1}\int_{1-y}^{1}f_{X}(x)dx \;f_Y(y)dy=\int_{\frac{1}{2}}^{1} y \, dy=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}.$ So since we know that $P(Y>1/2)=1/2$ you find: $$P(X+Y>1|Y>1/2)=\frac{3}{4}.$$