Given a vector of non-increasing positive integers $A=[a_1, \cdots, a_n]$, where $\sum_{1 \leq i \leq n}{a_i}=n(n-1)$, what is the probability that a random non-increasing vector $B=[b_1, \cdots, b_n]$, where $\sum_{1 \leq i \leq n}{b_i}=n(n-1)$, has the following property $\sum_{1 \leq i \leq n}{i \cdot a_i} = \sum_{1 \leq i \leq n}{i \cdot b_i}$?


That would depend on the distribution of $B$. However with a wide range of distributions it is almost certain (having zero probability) that $B$ does not have those properties. The sums will require $B$ to lie in an affine space of dimension $n-2$ which normally is a set of zero measure in $\mathbb R^n$. Even the prerequisite that $B$ is non-increasing should not alter this.

Otherwise there do exist such vectors. Take for example:

$$b_j = \begin{cases} \sum_{k=1}^n a_k & \text{if } j = 0 \\ 0 & \text{otherwise} \end{cases}$$

  • $\begingroup$ The condition is $\sum_{i} i \times b_i = \sum_{i} i \times a_i$, not just the sums of $a_i$ and $b_i$ must be equal. $\endgroup$ – Hetebrij Jan 9 '17 at 8:11
  • $\begingroup$ I phrased my question in the wrong way. I meant what is the probability of finding such vector $B$. Furthermore, as pointed out by @Hetebrij, the condition is different with yours. $\endgroup$ – orezvani Jan 9 '17 at 12:27
  • $\begingroup$ @emab I've updated the answer to reflect your changed question. Also note that probability of finding such vector and probability that a random vector is such is not the same thing (compare the probability of finding a six on a dice and actually getting a six when throwing the dice). $\endgroup$ – skyking Jan 9 '17 at 12:42

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