In each case the solution can be visualized as polygons within the rectangle bounded by 0< A <1 and 0< B <1. So we just need to identify the polygons and find their areas.
For X the areas are one big triangle for 1< X <2 and then a series of groups of trapezoids with their bases at 45 degrees.
For .9< X <1, .09< X <.1, .009< X <.01 etc the areas are 19/200, 19/20000, 19/2000000 etc
So the sum can be written as
19/200(1+1/100+1/1000...)
=19/200 * 100/99
=19/198
P(X=9)=19/198
by similar reasoning we find
P(X=8)=17/198
P(X=7)=15/198
P(X=6)=13/198
P(X=5)=11/198
P(X=4)=9/198
P(X=3)=7/198
P(X=2)=5/198
P(X=1)=3/198+1/2 = 102/198
Y is quite similar except there are two sets of trapezoids (with little triangles at the corners for .9) and the series of thin trapezoids approaches the diagonal A=B
For Y=1 the series has areas .17, .0197, .001997, etc which sums to .19191919...=19/99.
The full distribution is
P(Y=1)=19/99
P(Y=2)=17/99
P(Y=3)=15/99
P(Y=4)=13/99
P(Y=5)=11/99
P(Y=6)=9/99
P(Y=7)=7/99
P(Y=8)=5/99
P(Y=9)=3/99
Z is the most interesting. The shapes for each value are all triangles in two separate series.
For B>A the triangles are independent of Z. The areas for each are 1/20 + 1/200 + 1/2000 ... = 5/90.
P(1)=5/90+1/4+1/40+1/400...=5/90+5/18=1/3
P(2)=5/90+1/12+1/120+1/1200...=5/90+5/54=4/27
P(3)=5/90+1/24+1/240+1/2400...=5/90+5/108=11/108
P(4)=5/90+1/40+1/400+1/4000...=5/90+1/36=1/12
P(5)=5/90+1/60+1/600+1/6000...=5/90+1/54=2/27
P(6)=5/90+1/84+1/840+1/8400...=5/90+5/378=13/189
P(7)=5/90+1/112+1/1120+1/11200...=5/90+5/504=11/168
P(8)=5/90+1/144+1/1440+1/14400...=5/90+5/648=41/648
P(9)=5/90+1/180+1/1800+1/18000...=5/90+1/162=5/81 |