Problem 9.1 Easy Free

Conditional Region in the Unit Square

Let X,Y∼Unif(0,1)X,Y \sim \mathrm{Unif}(0,1) be independent. Find P[X−Y>1/2∣X+Y<1]\mathbb{P}[X - Y > 1/2 \mid X + Y < 1].

Work it out, then compare with the answer.

SolutionShow the worked solution

Conditioned on X+Y<1X+Y<1, the point (X,Y)(X,Y) is uniform on the triangle

T={(x,y):x,y≥0, x+y<1},T=\{(x,y): x,y\ge 0,\ x+y<1\},

which has area 12\tfrac12. The conditional probability is therefore the area of the part of TT satisfying x−y>12x-y>\tfrac12, divided by 12\tfrac12.

That part of TT is a triangle with the following sides:

  • it lies above y=0y=0;
  • it lies below the line y=x−12y=x-\tfrac12;
  • it lies below the line x+y=1x+y=1.

The two sloping lines meet where x−12=1−xx-\tfrac12=1-x. This gives the point (34,14)(\tfrac34,\tfrac14).

So the triangle has vertices (12,0)(\tfrac12,0), (1,0)(1,0) and (34,14)(\tfrac34,\tfrac14). Its base is 12\tfrac12 and its height is 14\tfrac14, so its area is

12⋅12⋅14=116.\tfrac12\cdot\tfrac12\cdot\tfrac14=\tfrac1{16}.

Hence

P ⁣[X−Y>12 | X+Y<1]=1/161/2=18.\mathbb{P}\!\left[X-Y>\tfrac12 \,\middle|\, X+Y<1\right]=\frac{1/16}{1/2}=\frac18 .
18\boxed{\tfrac18}