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YolandaQ · 2020年03月17日

问一道题:NO.PZ2020010303000013

问题如下:

If the kurtosis of some returns on a small-cap stock portfolio was 6, what would the degrees of freedom parameter be if they were generated by a generalized Student’s tυt_\upsilon? What if the kurtosis was 9?

选项:

解释:

In a Student’s t, the kurtosis depends only on the degree of freedom and is k = 3(v-2)/( v-4).

This can be solved so that k(v - 4) = 3(v - 2) so that kv - 4k = 3v - 6 and kv - 3v = 4k - 6.

Finally, solving for v, v =(4k-6)/ (k-3)

Plugging in v =(24-6)/(6-3)=18/3=6 and v =(36-6)/ (9-3)= 5.

The degree of freedom is v-1.

The kurtosis falls rapidly as v grows.

For example, if v = 12 then k = 3.75, which is only slightly higher than the kurtosis of a normal (3).

请问t分布的kurtosis算法需要记住吗?考试会考吗?

1 个答案

袁园_品职助教 · 2020年03月18日

同学你好!

最好要记住,老师上课也提过这道题。

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