Working with moments

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This post gives some background information on calculation involving moments.

Let X be a random variable. Let \mu=E(X) be its mean and \sigma^2=Var(X) be its variance. Thus \sigma is the standard deviation of X.

The expectation \displaystyle E[X^k] is the kth raw moment. It is also called the kth moment about zero. The expectation \displaystyle E[(X-\mu)^k] is the kth central moment. It is also called the kth moment about the mean. Given X, its standardized random variable is \displaystyle \frac{X-\mu}{\sigma}. Then the kth standardized moment is \displaystyle E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^k \biggr].

The mean of X is the first raw moment \mu=E(X). The variance of X is the second central moment \displaystyle Var(X)=E[(X-\mu)^2]. It is equivalent to Var(X)=E(X^2)-\mu^2. In words, the variance is the second raw moment subtracting the square of the mean.

The skewness of X is the third standardized moment of X. The kurtosis is the fourth standardized moment of X. The excess kurtosis is the kurtosis subtracting 3. They are:

    \displaystyle \gamma_1 =E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^3 \biggr]

    \displaystyle \text{kurt}[X] =E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^4 \biggr]

    \displaystyle \text{Ex Kurt}[X] = \text{kurt}[X]-3

The calculation is usually done by expanding the expression inside the expectation. As a result, the calculation would then be a function of the individual raw moments up to the third or fourth order.

    \displaystyle \begin{aligned} \gamma_1&=E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^3 \biggr] \\&=\frac{E(X^3)-3 \mu E(X^2)+3 \mu^2 E(X)-\mu^3}{(\sigma^2)^{1.5}} \\&=\frac{E(X^3)-3 \mu \sigma^2 - \mu^3}{(\sigma^2)^{1.5}} \end{aligned}

    \displaystyle \begin{aligned} \text{kurt}[X]&=E \biggl[\biggl(\frac{X-\mu}{\sigma} \biggr)^4 \biggr] \\&=\frac{E(X^4)-4 \mu E(X^3)+6 \mu^2 E(X^2)-4 \mu^3 E(X)+\mu^4}{\sigma^4} \\&=\frac{E(X^4)-4 \mu E(X^3)+6 \mu^2 E(X^2)-3 \mu^4}{\sigma^4} \end{aligned}

Note that the last line in skewness is a version that depends on the mean, the variance and the third raw moment. The calculation is illustrated with some examples. Practice problems are given in subsequent posts.

Example 1
Losses are modeled by a distribution that has the following density function. Calculate the mean, variance, skewness and kurtosis of the loss distribution.

    \displaystyle  f(x)=\left\{ \begin{array}{ll}                     \displaystyle  0.15 &\ 0<x \le 5 \\           \text{ } & \text{ } \\           \displaystyle  0.05 &\ 5<10           \end{array} \right.

The following shows the calculation of the first four raw moments.

    \displaystyle E(X)=\int_0^5 0.15 x \ dx+\int_5^{10} 0.05 x \ dx=3.75

    \displaystyle E(X^2)=\int_0^5 0.15 x^2 \ dx+\int_5^{10} 0.05 x^2 \ dx=\frac{62.5}{3}

    \displaystyle E(X^3)=\int_0^5 0.15 x^3 \ dx+\int_5^{10} 0.05 x^3 \ dx=140.625

    \displaystyle E(X^4)=\int_0^5 0.15 x^4 \ dx+\int_5^{10} 0.05 x^4 \ dx=1062.5

The following shows the results.

    \displaystyle Var(X)=\frac{62.5}{3}-3.75^2=\frac{20.3125}{3}=6.77083

    \displaystyle \gamma_1=\frac{140.625-3 \times 3.75 \times \displaystyle \frac{20.3125}{3} - 3.75^3}{\biggl(\displaystyle \frac{20.3125}{3} \biggr)^{1.5}}=0.66515

    \displaystyle \text{kurt}[X]=\frac{1062.5-4 \times 3.75 \times 140.625+ 6 \times 3.75^2 \times \displaystyle  \frac{62.5}{3}-3 \times 3.75^4}{\displaystyle  \biggl(\frac{20.3125}{3} \biggr)^2}=2.56686

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\copyright 2017 – Dan Ma

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3 thoughts on “Working with moments

    […] This post has two practice problems to complement the previous post Working with moments. […]

    […] moment (skewness) and the fourth standardized moment (kurtosis). To see how this is done, see this previous post. Another is to use the formulas given […]

    […] This post has two practice problems to complement the previous post Working with moments. […]

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