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

Using indicator random variables:

1. Let X = random variable which is equal to the number of inversions.
2. Let Xij = 1 if A[i] and A[j] form an inversion pair, and Xij = 0 otherwise.
3. Number of inversion pairs = Sum over 1 <= i < j <= n of (Xij)
4. Now P[Xij = 1] = P[A[i] > A[j]] = (n choose 2) / (2! * n choose 2) = 1/2
5. E[X] = E[sum over all ij pairs such that i < j of Xij] = sum over all ij pairs such that i < j of E[Xij] = n(n - 1) / 4

Answer 2

Another solution is even simpler, IMO, although it does not use "indicator random variables".

Since all of the numbers are distinct, every pair of elements is either an inversion (i < j with A[i] > A[j]) or a non-inversion (i < j with A[i] < A[j]). Put another way, every pair of numbers is either in order or out of order.

So for any given permutation, the total number of inversions plus non-inversions is just the total number of pairs, or n*(n-1)/2.

By symmetry of "less than" and "greater than", the expected number of inversions equals the expected number of non-inversions.

Since the expectation of their sum is n*(n-1)/2 (constant for all permutations), and they are equal, they are each half of that or n*(n-1)/4.

[Update 1]

Apparently my "symmetry of 'less than' and 'greater than'" statement requires some elaboration.

For any array of numbers A in the range 1 through n, define ~A as the array you get when you subtract each number from n+1. For example, if A is [2,3,1], then ~A is [2,1,3].

Now, observe that for any pair of numbers in A that are in order, the corresponding elements of ~A are out of order. (Easy to show because negating two numbers exchanges their ordering.) This mapping explicitly shows the symmetry (duality) between less-than and greater-than in this context.

So, for any A, the number of inversions equals the number of non-inversions in ~A. But for every possible A, there corresponds exactly one ~A; when the numbers are chosen uniformly, both A and ~A are equally likely. Therefore the expected number of inversions in A equals the expected number of inversions in ~A, because these expectations are being calculated over the exact same space.

Therefore the expected number of inversions in A equals the expected number of non-inversions. The sum of these expectations is the expectation of the sum, which is the constant n*(n-1)/2, or the total number of pairs.

[Update 2]

A simpler symmetry: For any array A of n elements, define ~A as the same elements but in reverse order. Associate the element at position i in A with the element at position n+1-i in ~A. (That is, associate each element with itself in the reversed array.)

Now any inversion in A is associated with a non-inversion in ~A, just as with the construction in Update 1 above. So the same argument applies: The number of inversions in A equals the number of inversions in ~A; both A and ~A are equally likely sequences; etc.

The point of the intuition here is that the "less than" and "greater than" operators are just mirror images of each other, which you can see either by negating the arguments (as in Update 1) or by swapping them (as in Update 2). So the expected number of inversions and non-inversions is the same, since you cannot tell whether you are looking at any particular array through a mirror or not.

Answer 3

Even simpler (similar to Aman's answer above, but perhaps clearer) ...

Let Xij be a random variable with Xij=1 if A[i] > A[j] and Xij=0 otherwise.
Let X=sum(Xij) over i, j where i < j

Number of pairs (ij)*:               n(n-1)/2
Probability that Xij=1 (Pr(Xij=1))): 1/2
By linearity of expectation**:       E(X) = E(sum(Xij))
                                          = sum(E(Xij))
                                          = sum(Pr(Xij=1))
                                          = n(n-1)/2 * 1/2
                                          = n(n-1)/4

*  I think of this as the size of the upper triangle of a square matrix.
** All sums here are over i, j, where i < j.

Answer 4

All the solutions seem to be correct, but the problem says that we should use indicator random variables. So here is my solution using the same:

    Let Eij be the event that i < j and A[i] > A[j].

    Let Xij = I{Eij} = {1 if (i, j) is an inversion of A

                        0 if (i, j) is not an inversion of A}

    Let X = Σ(i=1 to n)Σ(j=1 to n)(Xij) = No. of inversions of A.

    E[X] = E[Σ(i=1 to n)Σ(j=1 to n)(Xij)]

         = Σ(i=1 to n)Σ(j=1 to n)(E[Xij])

         = Σ(i=1 to n)Σ(j=1 to n)(P(Eij))

         = Σ(i=1 to n)Σ(j=i + 1 to n)(P(Eij)) (as we must have i < j)

         = Σ(i=1 to n)Σ(j=i + 1 to n)(1/2) (we can choose the two numbers in
                                            C(n, 2) ways and arrange them
                                            as required. So P(Eij) = C(n, 2) / n(n-1))

         = Σ(i=1 to n)((n - i)/2)

         = n(n - 1)/4

Answer 5

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