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粗眉毛辣椒油 · 2024年01月15日

这道题答案是在解释什么?

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).

根据公式T分布的kurtosis=3(n-2)/(n-4),df=n-1

两个问题分别为当K-6和5时,df为多少,不是依次代入公式算得df分别为5和4吗?


1 个答案

DD仔_品职助教 · 2024年01月15日

嗨,从没放弃的小努力你好:


同学你好,

是这样的。

这道题老师在基础班有讲过,同学如果不清楚可以再去回顾一下:

----------------------------------------------
努力的时光都是限量版,加油!

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