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🐳Yilemonee🐳 · 2025年06月09日

解释题目的意思和考点

NO.PZ2022062760000021

问题如下:

A portfolio manager holds five bonds in a portfolio and each bond has a 1-year default probability of 17%. The event of default for each of the bonds is independent. What is the mean and standard deviation of the number of bonds defaulting over the next year?

选项:

A.

Mean = 0.15, standard deviation = 0.71

B.

Mean = 0.85, standard deviation = 0.84

C.

Mean = 0.85, standard deviation = 0.71

D.

Mean = 0.15, standard deviation = 0.84

解释:

中文解析:

公式计算:

Mean = E(K) = n x p = 5 x 0.17 = 0.85.

Variance = Variance(K) = n x p x (1-p) = 5 x 0.17 x (0.83) = 0.7055

Standard deviation = sqrt(0.7055) = 0.8399.

选B

Letting n equal the number of bonds in the portfolio and p equal the individual default probability, the formulas to use are as follows:

Mean = E(K) = n x p = 5 x 0.17 = 0.85.

Variance = Variance(K) = n x p x (1-p) = 5 x 0.17 x (0.83) = 0.7055

Standard deviation = sqrt(0.7055) = 0.8399.

请解释题目的意思和考点

1 个答案

李坏_品职助教 · 2025年06月09日

嗨,努力学习的PZer你好:


题目说现在有五个债券,每个债券都是相互独立的,每个债券违约的概率是17%. 问你下一年违约的均值和标准差是多少?


既然每个债券相互独立,而每个债券在下一年只有两种情况,要么违约,要么不违约。这就类似于把一枚不规则的硬币连续掷五次,那就是典型的二项分布(Binomial distribution)。


题目问的就是这个二项分布的均值和标准差是多少。二项分布的均值 = n * p,n就是实验的次数(5个债券类似于掷硬币5次,n = 5),p是每次事件(违约)发生的概率,所以均值 = 5*0.17 = 0.85.


二项分布的方差 = n * p * (1-p) =  5 x 0.17 x (0.83) = 0.7055,那么标准差 = 根号0.7055 = 0.8399.

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