Hence find first four central moments

Solution :

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

By definition, moment generating function (M.G.F) about mean

$ \begin{aligned} M_x - (t) &=E[e^{t(x-\overline x)}] \\ &= \Sigma e^{t(x-\overline x)}p_i \\ &= \Sigma 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 \end {aligned} $

Now, central moments are given by

$$\mu_x=\Bigg[\dfrac {d^x}{dt^x}M_x-(t)\Bigg]_{t=0}$$

First Central Moment

$ \begin {aligned}{I} \mu_1^1 &=\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 \end{aligned} $

Second central moment

$ \begin {aligned}{I} \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 \end{aligned} $

Third central moment

$ \begin {aligned}{I} \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 \end{aligned} $

Fourth central moment

$ \begin {aligned}{I} \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 \end{aligned} $