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和棋 · 2020年08月20日

问一道题: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(xE[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[XE[X]]3/σ3=E[(Xμσ)3]=Σ(xμσ)3Pr(Xx)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[(XE[X])4]σ4=E[(Xμσ)4]=Σ(xμσ)4Pr(Xx)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%.

请问在c.d选项中Pr(X-x)是什么意思

1 个答案

小刘_品职助教 · 2020年08月21日

同学你好,

Pr(X-x) 是Pr(X)的概率,与X-x 相乘的意思,这个第三四问不需要掌握计算,考试的时候只需要会判断左偏还是右偏 尖峰还是矮峰就行了。

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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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