考拉 · 2024年01月20日
* 问题详情，请 查看题干
NO.PZ202207040100000604
问题如下：
As an indexing technique, the number of holdings in Manager B’s index least likely illustrates:
选项：
A.Reconstitution. B.packeting. C.buffering.解释：
Solution
Reconstitution
involves deleting names that are no longer in the index and adding names that
have been approved as new index members. Therefore the holdings of the fund
should always be 500.
请问为什么在packeting方法下，portfolio的股票只会比benchmark多或者相等，而buffering 下，持股多少都可以
笛子_品职助教 · 2024年01月22日
嗨，爱思考的PZer你好：
请问为什么在packeting方法下，portfolio的股票只会比benchmark多或者相等，而buffering 下，持股多少都可以
Hello，亲爱的同学~
我们看Packeting的做法。
当benchmark移除一只股票，并加入一只股票时，portfolio只会移除这只股票的一部分，然后加入新股票的一部分。
但是在计算股票的number的时候，我们只会算是增加了一只股票，或者减少了一只股票。
举例来说：
A股票占benchmark的权重为3%，benchmark共有30只股票。
portfolio复制benchmark，A股票也占portfolio的权重5%，portfolio也有30只股票。
之后，benchmark做持仓调整，剔除了A股票，调入了B股票。B股票完全取代A股票，即B股票权重也是3%。此时portfolio减少1只A，加入B，股票数量还是100只。
portfolio由于采用packeting方法，剔除A股票的时候，只剔除了1%权重，还保留2%权重，因此，portfolio还有A股票。
同样的。B股票，portfolio调入了1%权重，尽管没有像benchmark一样调入3%，权重，但1%的持股也意味着portfolio持有B股票。
因此，benchmark没有A只有B，portfolio既有A又有B。portfolio的股票数量是101只。
因此，packeting方法下，portfolio股票数量 大于等于 benchmark股票数量，不会小于。
----------------------------------------------虽然现在很辛苦，但努力过的感觉真的很好，加油！
NO.PZ202207040100000604问题如下 inxing technique, the number of holngs in Manager B’s inx least likely illustrates: A.Reconstitution.B.packeting.C.buffering. SolutionReconstitutioninvolves leting names thare no longer in the inx anaing names thathave been approvenew inx members. Therefore the holngs of the funhoulalways 500. 从哪里看出来always 500?
NO.PZ202207040100000604问题如下 inxing technique, the number of holngs in Manager B’s inx least likely illustrates: A.Reconstitution.B.packeting.C.buffering. SolutionReconstitutioninvolves leting names thare no longer in the inx anaing names thathave been approvenew inx members. Therefore the holngs of the funhoulalways 500. 看了老师对其他问题的解答，理解了为什么packeting的方法会使得portfolio的股票数量只会大于或等于benchmark的股票数量，希望再进一步解答为什么buffer的方法会让股票数量大于，等于，小于都可以呢。知道buffer是设置一个安全垫，所以大于或者等于benchmark的股票数量能理解，只不过小于的话是怎么实现的呢，剔除掉组合中股票的一部分？不还是算一只么，谢谢。
NO.PZ202207040100000604 问题如下 inxing technique, the number of holngs in Manager B’s inx least likely illustrates: A.Reconstitution. B.packeting. C.buffering. SolutionReconstitutioninvolves leting names thare no longer in the inx anaing names thathave been approvenew inx members. Therefore the holngs of the funhoulalways 500. RT
NO.PZ202207040100000604 问题如下 inxing technique, the number of holngs in Manager B’s inx most likely illustrates: A.reconstituting. B.packeting. C.buffering. SolutionC is correct. Buffering involves establishing ranges arounbreakpoints thfine whether a stobelongs in one inx or another. Some inx provirs have aptepolicies intento limit stomigration problems ankeep trang costs low for investors who replicate inxes. Size rankings mchange ily with market primovements, so buffering makes inx transitions a more graanorrly process. long stocks remain within the buffer zone, they stin the current inx, ana result, the holngs of the funmexceethe holngs of the inx.A is incorrect. Reconstitution involves leting names thare no longer in the inx anaing names thhave been approvenew inx members. B is incorrect. Packeting involves splitting stopositions into multiple parts. For example, if a micstock’s capitalization increases anbreaches the breakpoint between the micanlarge-cinxes, a portion of the totholng is transferreto the large-cinx but the rest stays in the micinx. On the next reconstitution te if the stovalue remains large canall other qualifications are met, the remainr of the shares are moveout of the micinx into the large-cinx. 一个月了这道题答案还不修改？？？？？？？？？？？？？？？？
