x and y are two real numbers, each of which is uniformly randomly chosen in the interval (1,5).
Determine the probability of each of the following events:
(i) ⌈3*x⌉ + ⌈4*y⌉ = ⌈3*x + 4*y⌉
(ii) ⌈3*x⌉ - ⌈4*y⌉ = ⌈3*x - 4*y⌉
(iii) ⌈3*x⌉ * ⌈4*y⌉ = ⌈12*x*y⌉
⌈3*x⌉
(iv) --------- = ⌈(3*x)/(4*y)⌉
⌈4*y⌉
*** ⌈n⌉ denotes ceiling(n), that is, the lowest integer greater than or equal to n.
Problems (i) and (ii) look complicated but they are each the same
problem posed over and over, 12x16 times each. (i) asks if the sum
of two uniformly distributed random variables,
when added, add with enough in the decimal places
to make the addition have a carry to the ones place e.g:
1.2 + 1.6 rounds (using ceiling) to 2 while 1.2 + 1.9 rounds to 3.
The sum of two Boxcar uniform PDFs convolve to make a Triangular with a
PDF symmetric around the sum. So, half the results round up, the other half don't.
There are ample examples of the math for this online where the
boxcars simply convolve make a triangle.
(iii) is where I have made progress. Each product is (j+del) * (k+del)
The new random variables are made from x = (j to j+1) and y = (k to k+1) and
the required probability is calculated by
Pr{ Ceil(X)Ceil(Y) = Ceil(XY) }
= int{ j to (jk+1)/k } int{k to (jk+1)/x} dy dx
= (jk+1) [ ln( (jk+1)/k ) - ln(j) ] - 1
There are all the j=1,15 and k=1,20 pairs. All j,k pairs
appear twice (e.g. 3,4 4,3) when both are le 15 and j ne k.
These j, k pairs should sum with their individual
probability when weighted for appearing twice,
to the overall probability. TBC...
Edited on December 7, 2022, 3:47 pm
progress on the math
