The expected number of inversions--From Introduction to Algorithms by Cormen

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 inv

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