Answer to Question #242046 in Statistics and Probability for ananya

Question

Find the variance of the r.v. whose m.g.f is (𝑒^−𝑡)/12(2+𝑒^𝑡+6𝑒^3𝑡+3𝑒^6𝑡)

Answer 1

Solution:

$M(t)=\dfrac{𝑒^{−𝑡}}{12(2+𝑒^𝑡+6𝑒^{3𝑡}+3𝑒^{6𝑡})}$

$=\dfrac{1}{12(2𝑒^{𝑡}+𝑒^{2𝑡}+6𝑒^{4𝑡}+3𝑒^{7𝑡})} \\=\dfrac{1}{12}(2𝑒^{𝑡}+𝑒^{2𝑡}+6𝑒^{4𝑡}+3𝑒^{7𝑡})^{-1}$

On differentiating w.r.t $t$,

$M'(t)=\dfrac{-1}{12}(2𝑒^{𝑡}+𝑒^{2𝑡}+6𝑒^{4𝑡}+3𝑒^{7𝑡})^{-2}(2𝑒^{𝑡}+2𝑒^{2𝑡}+24𝑒^{4𝑡}+21𝑒^{7𝑡})$

$=-\dfrac{21\mathrm{e}^{7t}+24\mathrm{e}^{4t}+2\mathrm{e}^{2t}+2\mathrm{e}^t}{12\left(3\mathrm{e}^{7t}+6\mathrm{e}^{4t}+\mathrm{e}^{2t}+2\mathrm{e}^t\right)^2}$

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$=0.08667$