Hence find first four central moments.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 04

Year : MAY 2015

By definition, Mean $\overline x = E(x)\\ =\sum x_ip_i\\ =-2.\dfrac 13+3.\dfrac 12+1.\dfrac 16\\ =1$

By definition, moment generating function (MGf) about mean $M_x - (t)=E[e^{t(x-\overline x)}]\\ =\sum e^{t(x-\overline x)}p_i\\ =\sum e^{t(x-1)}p_i\\ =e^{-3t}.\dfrac 13+2^{2t}.\dfrac 12+e^{0t}.\dfrac 16\\ =\dfrac 13e^{-3t}+\dfrac 12e^{2t}+\dfrac 16$

Now, central moments are given by $\mu_x=\Bigg[\dfrac {d^x}{dt^x}M_x-(t)\Bigg]_{t=0}$

First central moment $\mu_1=\Bigg[\dfrac d{dt}M_x (t)\Bigg]_{t=0}\\ =\Bigg[\dfrac d{dt}\Bigg(\dfrac 13e^{-3t}+\dfrac 12e^{2t}+\dfrac 16\Bigg)\Bigg]_{t=0}\\ =\Bigg[\dfrac 13e^{-3t}.(-3)+\dfrac 13e^{2t}.2+0\Bigg]_{t=0}......(1)\\ =-1+1+0\\ =0$

Second central moment $\mu_2^1=\Bigg[\dfrac {d^2}{dt^2}M_x-(t)\Bigg]_{t=0}\\ =\Bigg[\dfrac {d^2}{dt^2}\Bigg(\dfrac 13e^{-3t}+\dfrac 12e^{2t}+\dfrac 16\Bigg)\Bigg]_{t=0} \\ = \Bigg[\dfrac d{dt}(-e^{-3t}+e^{2t})\Bigg]_{t=0} \space \space \space \space \space from (1)\\ =[-e^{-3t}.(-3)+e^{2t}.2]_{t=0}......(2)\\ =3+2=5$

Third central moment $\mu_3^1=\Bigg[\dfrac {d^3}{dt^3}M_x -(t)\Bigg]_{t=0}\\ =\Bigg[\dfrac {d^3}{dt^3}\Bigg(\dfrac 13e^{-3t}+\dfrac 12e^{2t}+\dfrac 16\Bigg)\Bigg]_{t=0}\\ = \Bigg[\dfrac d{dt}(3e^{-3t}+2e^{2t})\Bigg]_{t=0} \space \space \space \space \space from (2)\\ =[3e^{-3t}.(-3)+2e^{2t}.2]_{t=0}......(3)\\ =-9+4\\ =5$

Fourth central moment $\mu_4^1=\Bigg[\dfrac {d^4}{dt^4}M_x -(t)\Bigg]_{t=0}\\ =\Bigg[\dfrac {d^4}{dt^4}\Bigg(\dfrac 13e^{-3t}+\dfrac 12e^{2t}+\dfrac 16\Bigg)\Bigg]_{t=0}\\ = \Bigg[\dfrac d{dt}(-9e^{-3t}+4e^{2t})\Bigg]_{t=0} \space \space \space \space \space from (3)\\ =[-9e^{-3t}.(-3)+4e^{2t}.2]_{t=0} \\ =27+8\\ =35$