Answer to Question #234694 in Statistics and Probability for dhruv

Question

The first four moments of a distribution about the value ‘4’ of the variable are -1.5,17,-30 and 108. Find the moments about mean and calculate the coefficient of skewness and kurtosis

Answer 1

Solution:

$β_n =M(ξ−M_ξ)\ —\ the\ n-th\ central\ moment. \\1)M(ξ−4)=−1.5 \\Mξ=4−1.5=2.5 \\M(ξ−Mξ)=0\ —\ the\ first\ central\ moment.$

$2) M(\xi-4)^{2}=17 \\M(\xi-M \xi)^{2}=M \xi^{2}-(M \xi)^{2} \\M \xi^{2}-4^{2}=17 \\M \xi^{2}=33$

$\begin{aligned} &M(\xi-M \xi)^{2}=33-(2.5)^{2}=26.75-\text { the second central moment. } \\ &\text { 3) } M(\xi-4)^{3}=-30 \\ &M(\xi-M \xi)^{3}=M\left(\xi^{3}-3 \xi^{2} M \xi+3 \xi(M \xi)^{2}-(M \xi)^{3}\right)=M \xi^{3}-3 M \xi^{2} M \xi+ \\ &3(M \xi)^{3}-(M \xi)^{3}=M \xi^{3}-3 M \xi^{2} M \xi+2(M \xi)^{3} \\ &M(\xi-4)^{3}=M \xi^{3}-3 \cdot 33 \cdot 4+2 \cdot 4^{3}=-30 \\ &M \xi^{3}=238 \end{aligned}$

$M(\xi-M \xi)^{3}=238-3 \cdot 33 \cdot 2.5+2 \cdot(2.5)^{3}=21.75-\text{ the third central moment}$

$4) M(\xi-4)^{4}=108$

$\begin{aligned} &M(\xi-M \xi)^{4}=M\left(\xi^{4}-4 \xi^{3} M \xi+6 \xi^{2}(M \xi)^{2}-4 \xi(M \xi)^{3}+(M \xi)^{4}\right)= \\ &=M \xi^{4}-4 M \xi^{3} M \xi+6 M \xi^{2}(M \xi)^{2}-4(M \xi)^{4}+(M \xi)^{4}= \\ &=M \xi^{4}-4 M \xi^{3} M \xi+6 M \xi^{2}(M \xi)^{2}-3(M \xi)^{4} \\ &M(\xi-4)^{4}=M \xi^{4}-4 \cdot 238 \cdot 4+6 \cdot 33 \cdot 4^{2}-3 \cdot 4^{4}=108 \end{aligned}$

$\begin{aligned} &M \xi^{4}=1516 \\ &M(\xi-2.5)^{4}=1516-4 \cdot 238 \cdot 2.5+6 \cdot 33 \cdot(2.5)^{2}-3 \cdot(2.5)^{4}= \end{aligned}$

$256.3125 - \text{the fourth central moment}. \\\Rightarrow\sigma \approx 5.17$

$A=\frac{M(\xi-M \xi)^{3}}{\sigma^{3}}=\frac{21.75}{138.35}=0.157- \text{asymmetry of the distribution}. \\E=\frac{M(\xi-M \xi)^{4}}{\sigma^{4}}=\frac{256.3125}{715.5625}=0.358- \text{kurtosis of the distribution}.$