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, 1 β and 2 β . Find also the moments about (i) the origin, and (ii) the point x=2.

Answer 1

$i) \beta_n=M(\xi-M\xi)\text{ --- the n-th central moment}.\\ 1) M(\xi-4)=-1.5\\ M\xi=4-1.5=2.5\\ M(\xi-M\xi)=0\text{ --- the first cental moment}.\\ 2) M(\xi-4)^2=17\\ M(\xi-M\xi)^2=M\xi^2-(M\xi)^2\\ M\xi^2-4^2=17\\ M\xi^2=33\\ M(\xi-M\xi)^2=33-(2.5)^2=26.75\text{ --- the second central moment}.\\ 3) M(\xi-4)^3=-30\\ M(\xi-M\xi)^3=M(\xi^3-3\xi^2M\xi+3\xi(M\xi)^2-(M\xi)^3)=M\xi^3-3M\xi^2M\xi+3(M\xi)^3-(M\xi)^3=M\xi^3-3M\xi^2M\xi+2(M\xi)^3.\\ M(\xi-4)^3=M\xi^3-3\cdot 33\cdot 4+2\cdot 4^3=-30\\ M\xi^3=238.\\ M(\xi-M\xi)^3=238-3\cdot 33\cdot 2.5+2\cdot (2.5)^3=21.75\text{ --- the third central moment}.\\ 4) M(\xi-4)^4=108.\\ M(\xi-M\xi)^4=M(\xi^4-4\xi^3M\xi+6\xi^2(M\xi)^2-4\xi(M\xi)^3+(M\xi)^4)=\\ =M\xi^4-4M\xi^3M\xi+6M\xi^2(M\xi)^2-4(M\xi)^4+(M\xi)^4=\\ =M\xi^4-4M\xi^3M\xi+6M\xi^2(M\xi)^2-3(M\xi)^4.\\ M(\xi-4)^4=M\xi^4-4\cdot 238\cdot 4+6\cdot 33\cdot 4^2-3\cdot 4^4=108.\\ M\xi^4=1516.\\ M(\xi-2.5)^4=1516-4\cdot 238\cdot 2.5+6\cdot 33\cdot (2.5)^2-3\cdot(2.5)^4=256.3125\text{ --- the fourth central moment}.\\ \sigma\approx 5.17.\\ A=\frac{M(ξ−Mξ)^3}{\sigma^3}​=\frac{21.75}{138.35}=0.157 — \text{ asymmetry of the distribution}.\\ E=\frac{M(ξ−Mξ)^4​}{\sigma^4}=\frac{256.3125}{715.5625}​=0.358 — \text{ kurtosis of the distribution}.\\ \text{Moments about the origin:}\\ M\xi=2.5\\ M\xi^2=33\\ M\xi^3=238\\ M\xi^4=1516.$