让A [1 .. N]为N 不同的数字数组。如果I＆LT; j和A [1]> A [J]，则对（I，J）被称为A的逆（见问题2-4更多的倒置。）假设A的每个元素是随机选择，独立并均匀从范围1至n。使用指标随机变量来计算反转的预期数量。

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.

现在的问题是，从运动5.2-5中的算法导论的由Cormen。这是我的递归解决方案：

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

假定x（i）是反转的在[1..i]，和E（ⅰ）的数量为x的（ⅰ）的预期值，则E（i + 1的）可以计算如下：   图片我们有 I + 1 位置把所有的数字，如果我们把的 I + 1 的第一个位置，那么x（I + 1 ）= I + X（ⅰ）;如果我们把的 i + 1的的在第二位置，则x第（i + 1）= I-1 + X（i）中，...，所以E（i + 1的）= 1 /（ I + 1）*和（k）的+ E（i）中，其中，k = [0,1]。最终我们得到E（I + 1）= I / 2 + E（I）。   因为我们知道，E（2）= 0.5，所以递归我们得到：E（N）=（N-1 + N-2 + ... + 2）/ 2 + 0.5 = N *（N-1）/ 4

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.

虽然扣除上述似乎是正确的，但我仍然不是很确信这一点。所以，我在这里分享。

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

如果有什么不对，请大家指正。

If there is something wrong, please correct me.

我认为这是正确的，但我认为正确的方式来证明它是使用conditionnal预期：

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

对于所有的X和Y，我们有：E [X] = E [E [X | Y]]

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

然后在您的情况：

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）

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)

有关第二条语句：

如果：

E（N）= N *（N-1）/ 4。

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

则E（N + 1）=（N + 1）* N / 4 =（N-1）* N / 4 + 2 * N / 4 =（N-1）* N / 4 + N / 2 = E（N）+ N / 2

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

所以，N *（N-1）/ 4。验证对所有的n> = 2的递推关系，并验证其在n = 2

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

那么E（N）= N *（N-1）/ 4

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

希望我理解你的问题，它可以帮助

Hope I understood your problem and it helps