X and Y are two independent uniformly distributed random variables in [0; 1]. Then
P(Y-X<=1/2) is ??
Now from a graphical point of view, I plot the unit side square [0,1], and the line y= x+1/2, since I want to find the area which is below the triangle, I do 1- area(triangle) = 1- 1/2*1/2*1/2 = 7/8
and that's the correct answer
But I want to find that probability applying a general method, using the density function and making a double integral. They are independent
so I have to
\[\int_{0}^{1}\int_{0}^{1/2+x} fxfy dy dx\]=\[\int_{0}^{1}\int_{0}^{1/2+x} 1 dy dx\]
because fxfy =fx*fy= 1*1=1
But I get 1, so I make mistakes in defining the extrema
P(Y-X<=1/2) is ??
Now from a graphical point of view, I plot the unit side square [0,1], and the line y= x+1/2, since I want to find the area which is below the triangle, I do 1- area(triangle) = 1- 1/2*1/2*1/2 = 7/8
and that's the correct answer
But I want to find that probability applying a general method, using the density function and making a double integral. They are independent
so I have to
\[\int_{0}^{1}\int_{0}^{1/2+x} fxfy dy dx\]=\[\int_{0}^{1}\int_{0}^{1/2+x} 1 dy dx\]
because fxfy =fx*fy= 1*1=1
But I get 1, so I make mistakes in defining the extrema