Joint distrubtion question

[mnf]

if the joint p.d.f of x,y is given by

f(x,y)= a |x-y| ,0<=x,y<=1
f(x,y)= 0 , o.w


find i)a
ii)p(y>x)
iii)p(x=y)

 

[HallsofIvy]

if the joint p.d.f of x,y is given by

f(x,y)= a |x-y| ,0<=x,y<=1
f(x,y)= 0 , o.w


find i)a

a must be such that the total probability is 1. $x\ge y$, and so |x-y|= x- y, for (x,y) on the triangle with vertices at (0,0), (1, 1), and (1, 0). $y\ge x$, and so |x- y|= y- x for (x,y) on the triangle with vertices at (0, 0), (1, 1), and (0, 1). That is, we must have
$$\int_{x=0}^1\int_{y= 0}^x a(y- x) dydx+ \int_{x=0}^1\int_{y=x}^1 a(x- y)dydx= 1$$
$$a\left(\int_{x=0}^1\int_{y= 0}^x (y- x) dydx+ \int_{x=0}^1\int_{y= x}^1 (x- y)dydx\right)= 1$$
Do the integration on the left and take its reciprocal to find a.

ii)p(y>x)

Again, y> x in the triangle with vertices at (0,0), (1, 1), and (, 1).
$$p(y>x)= a\int_{x=0}^1\int_{y=x}^1 (y- x) dy dx$$
where "a" is the value from (i). (Though the answer should be obvious from symmetry.)

iii)p(x=y)

Any double integral over a line is 0.