Stats problem involving double integers

**lllll** · April 6th 2008, 03:46 PM

$f(x,y) = \left\{ \begin{array}{rcl}<br /> 1 & \mbox{for} & 0 \leq x \leq 1, \ \ 0 \leq y \leq 1\\<br /> 0 & \mbox{for} & \mbox{otherwise}<br /> \end{array}\right.$

What is $P \left( x-y > \frac{1}{2} \right)$

would the solution be $\int^{0.5}_{0} \int^{0.5-y}_0 1 dxdy = \ \ \int^{0.5}_{0} x \ \bigg{|} ^{0.5-y}_{0} dy = \ \ \int^{0.5}_{0} 0.5-y \ dy = \ \ 0.25y -\frac{y^2}{2} \ \bigg{|}^{0.5}_{0}$ ?

**falcald** · April 6th 2008, 04:14 PM

If 0<=x<=1 and 0<=y<=1 and taking into account that x-y>1/2 can be expressed as y<x-1/2, then the intervals in which you have to integrate the density functions are 1/2<=x<=1 and 0<=y<=x-1/2, so... your answer is not correct.

In your answer, check x=0 and y=1/2... it does not satisfy x-y>1/2 because 0-1/2<1/2.

It is a good practice to graph the intervals and draw the line which represents x-y=1/2. Then, you paint the area for x-y>1/2.

Hope this helps.

Regards,

Federico.

Federico.

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