Find the variance of the r.v. whose m.g.f is (π^βπ‘)/12(2+π^π‘+6π^3π‘+3π^6π‘)
Solution:
"M(t)=\\dfrac{\ud835\udc52^{\u2212\ud835\udc61}}{12(2+\ud835\udc52^\ud835\udc61+6\ud835\udc52^{3\ud835\udc61}+3\ud835\udc52^{6\ud835\udc61})}"
"=\\dfrac{1}{12(2\ud835\udc52^{\ud835\udc61}+\ud835\udc52^{2\ud835\udc61}+6\ud835\udc52^{4\ud835\udc61}+3\ud835\udc52^{7\ud835\udc61})}\n\\\\=\\dfrac{1}{12}(2\ud835\udc52^{\ud835\udc61}+\ud835\udc52^{2\ud835\udc61}+6\ud835\udc52^{4\ud835\udc61}+3\ud835\udc52^{7\ud835\udc61})^{-1}"
On differentiating w.r.t "t",
"M'(t)=\\dfrac{-1}{12}(2\ud835\udc52^{\ud835\udc61}+\ud835\udc52^{2\ud835\udc61}+6\ud835\udc52^{4\ud835\udc61}+3\ud835\udc52^{7\ud835\udc61})^{-2}(2\ud835\udc52^{\ud835\udc61}+2\ud835\udc52^{2\ud835\udc61}+24\ud835\udc52^{4\ud835\udc61}+21\ud835\udc52^{7\ud835\udc61})"
"=-\\dfrac{21\\mathrm{e}^{7t}+24\\mathrm{e}^{4t}+2\\mathrm{e}^{2t}+2\\mathrm{e}^t}{12\\left(3\\mathrm{e}^{7t}+6\\mathrm{e}^{4t}+\\mathrm{e}^{2t}+2\\mathrm{e}^t\\right)^2}"
Again, differentiating,
"M''(t)=\\dfrac{\\mathrm{e}^{-t}\\left(441\\mathrm{e}^{12t}+846\\mathrm{e}^{9t}+9\\mathrm{e}^{7t}+444\\mathrm{e}^{6t}+72\\mathrm{e}^{4t}-12\\mathrm{e}^{3t}+4\\mathrm{e}^{2t}+6\\mathrm{e}^t+4\\right)}{12\\left(3\\mathrm{e}^{6t}+6\\mathrm{e}^{3t}+\\mathrm{e}^t+2\\right)^3}"
Now, "M'(0)=-\\dfrac{21\\mathrm{e}^{0}+24\\mathrm{e}^{0}+2\\mathrm{e}^{0}+2\\mathrm{e}^0}{12\\left(3\\mathrm{e}^{0}+6\\mathrm{e}^{0}+\\mathrm{e}^{0}+2\\mathrm{e}^0\\right)^2}=-\\dfrac{49}{1728}"
And "M''(0)=\\dfrac{441+846+9+444+72-12+4+6+4}{12(3+6+1+2)^3}=\\dfrac{907}{10368}"
Then, variance"=M''(0)-[M'(0)]^2=\\dfrac{907}{10368}-(-\\dfrac{49}{1728})^2"
So, variance"=0.08667"
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