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yang · 2021年03月16日

请问用计算器如何计算出这组数据的平均数、方差等，我按了之后出现error，是因为输入了%号吗，请老师给出按键过程

  1. FRM I
  2. Quantitative

NO.PZ2020010302000010

问题如下：

Suppose the return on an asset has the following distribution:

a. Compute the mean, variance, and standard deviation.

b. Verify your result in (a) by computing E[X2]E[X^2] directly and using the alternative expression for the variance.

c. Is this distribution skewed?

d. Does this distribution have excess kurtosis? e. What is the median of this distribution?

选项：

解释：

a. The mean is E[X] = Σx Pr(X = x) = 0.25%.

The variance is Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555Var[X] = Σ(x - E[X])^2 Pr(X = x) = 0.000555.

The standard deviation is Var[X]=2.355\sqrt {Var[X]} = 2.355%.

b. E[X2]=Σx2Pr(X=x)=.000561E[X^2] = Σx^2 Pr(X = x) = .000561 and so E[X2]−(E[X])2=0.000561−(.0025)2=.000555E[X^2] - (E[X])^2 = 0.000561 - (.0025)^2 = .000555, which is the same.

c. The skewness requires computing

skew(X)=E[X−E[X]]3/σ3=E[(X−μσ)3]=Σ(x−μσ)3Pr(X−x)skew(X)=E[X-E[X]]^3/{\sigma^3}=E[(\frac{X-\mu}{\sigma})^3]=Σ(\frac{x-\mu}{\sigma})^3Pr(X-x)

Thus the skewness is 0.021, and the distribution has a mild positive skew.

d. The kurtosis requires computing

kurtosis(X)=E[(X−E[X])4]σ4=E[(X−μσ)4]=Σ(x−μσ)4Pr(X−x)kurtosis(X)=\frac{E[(X-E[X])^4]}{\sigma^4}=E[(\frac{X-\mu}{\sigma})^4]=Σ(\frac{x-\mu}{\sigma})^4Pr(X-x)

Thus the kurtosis is 2.24. The excess kurtosis is then 2.24 - 3 = -0.76. This distribution does not have excess kurtosis.

e. The median is the value where at least 50% probability lies to the left, and at least 50% probability lies to the right. Cumulating the probabilities into a CDF, this occurs at the return value of 0%.



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袁园_品职助教 · 2021年03月17日

同学你好！

计算器的使用可以去试听课程里面听一下，不过我记得计算器好像只能算equally weighted的一组数

这题给的就是数值和他们对应的概率，可以直接套公式求出来期望和标准差

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NO.PZ2020010302000010问题如下Suppose the return on asset hthe following stribution:Compute the mean, variance, anstanrviation.Verify your result in (computing E[X2]E[X^2]E[X2] rectly anusing the alternative expression for the variance.Is this stribution skewe es this stribution have excess kurtosis? e. Whis the meof this stribution?The meis E[X] = Σx Pr(X = x) = 0.25%.The varianis Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555Var[X] = Σ(x - E[X])^2 Pr(X = x) = 0.000555Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555.The stanrviation is Var[X]=2.355\sqrt {Var[X]} = 2.355%Var[X]​=2.355.E[X2]=Σx2Pr(X=x)=.000561E[X^2] = Σx^2 Pr(X = x) = .000561E[X2]=Σx2Pr(X=x)=.000561 anso E[X2]−(E[X])2=0.000561−(.0025)2=.000555E[X^2] - (E[X])^2 = 0.000561 - (.0025)^2 = .000555E[X2]−(E[X])2=0.000561−(.0025)2=.000555, whiis the same.The skewness requires computingskew(X)=E[X−E[X]]3/σ3=E[(X−μσ)3]=Σ(x−μσ)3Pr(X−x)skew(X)=E[X-E[X]]^3/{\sigma^3}=E[(\frac{X-\mu}{\sigma})^3]=Σ(\frac{x-\mu}{\sigma})^3Pr(X-x)skew(X)=E[X−E[X]]3/σ3=E[(σX−μ​)3]=Σ(σx−μ​)3Pr(X−x)Thus the skewness is 0.021, anthe stribution ha milpositive skew. The kurtosis requires computingkurtosis(X)=E[(X−E[X])4]σ4=E[(X−μσ)4]=Σ(x−μσ)4Pr(X−x)kurtosis(X)=\frac{E[(X-E[X])^4]}{\sigma^4}=E[(\frac{X-\mu}{\sigma})^4]=Σ(\frac{x-\mu}{\sigma})^4Pr(X-x)kurtosis(X)=σ4E[(X−E[X])4]​=E[(σX−μ​)4]=Σ(σx−μ​)4Pr(X−x)Thus the kurtosis is 2.24. The excess kurtosis is then 2.24 - 3 = -0.76. This stribution es not have excess kurtosis.e. The meis the value where least 50% probability lies to the left, anleast 50% probability lies to the right. Cumulating the probabilities into a C, this occurs the return value of 0%.1、老师，第一个和第二个红框是啥意思啊？2、为啥当X=0时，累积概率是52%，中位数不用线性插值法或者求X=0和X=-1的平均数呢？考试如果也是离散分布，累计概率超过50%，中位数也是取能概率一直累积超过50%的那个数？

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2024-01-30 12:48 2 · 回答