Question

Let A[1 .. n] be an array of n distinct numbers. If i < j and A[i] > A[j], then the pair (i, j) is called an inversion of A. (See Problem 2-4 for more on inversions.) Suppose that each element of A is chosen randomly, independently, and uniformly from the range 1 through n. Use indicator random variables to compute the expected number of inversions.

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The problem is from exercise 5.2-5 in Introduction to Algorithms by Cormen. Here is my recursive solution:

Suppose x(i) is the number of inversions in a[1..i], and E(i) is the expected value of x(i), then E(i+1) can be computed as following:
Image we have i+1 positions to place all the numbers, if we place i+1 on the first position, then x(i+1) = i + x(i); if we place i+1 on the second position, then x(i+1) = i-1 + x(i),…, so E(i+1) = 1/(i+1)* sum(k) + E(i), where k = [0,i]. Finally we get E(i+1) = i/2 + E(i).
Because we know that E(2) = 0.5, so recursively we get: E(n) = (n-1 + n-2 + … + 2)/2 + 0.5 = n* (n-1)/4.

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Although the deduction above seems to be right, but I am still not very sure of that. So I share it here.

If there is something wrong, please correct me.

Answer 1

I think it’s right, but I think the proper way to prove it is to use conditionnal expectations :

for all X and Y we have : E[X] =E [E [X|Y]]

then in your case :

E(i+1) = E[x(i+1)] = E[E[x(i+1) | x(i)]] = E[SUM(k)/(1+i) + x(i)] = i/2 + E[x(i)] = i/2 + E(i)

about the second statement :

if :

E(n) = n* (n-1)/4.

then E(n+1) = (n+1)*n/4 = (n-1)*n/4 + 2*n/4 = (n-1)*n/4 + n/2 = E(n) +n/2

So n* (n-1)/4. verify the recursion relation for all n >=2 and it verifies it for n=2

So E(n) = n*(n-1)/4

Hope I understood your problem and it helps