Answer to Question #242046 in Statistics and Probability for ananya

Question #242046

Find the variance of the r.v. whose m.g.f is (𝑒^βˆ’π‘‘)/12(2+𝑒^𝑑+6𝑒^3𝑑+3𝑒^6𝑑)


1
Expert's answer
2021-09-27T16:46:27-0400

Solution:

M(t)=π‘’βˆ’π‘‘12(2+𝑒𝑑+6𝑒3𝑑+3𝑒6𝑑)M(t)=\dfrac{𝑒^{βˆ’π‘‘}}{12(2+𝑒^𝑑+6𝑒^{3𝑑}+3𝑒^{6𝑑})}

=112(2𝑒𝑑+𝑒2𝑑+6𝑒4𝑑+3𝑒7𝑑)=112(2𝑒𝑑+𝑒2𝑑+6𝑒4𝑑+3𝑒7𝑑)βˆ’1=\dfrac{1}{12(2𝑒^{𝑑}+𝑒^{2𝑑}+6𝑒^{4𝑑}+3𝑒^{7𝑑})} \\=\dfrac{1}{12}(2𝑒^{𝑑}+𝑒^{2𝑑}+6𝑒^{4𝑑}+3𝑒^{7𝑑})^{-1}

On differentiating w.r.t tt,

Mβ€²(t)=βˆ’112(2𝑒𝑑+𝑒2𝑑+6𝑒4𝑑+3𝑒7𝑑)βˆ’2(2𝑒𝑑+2𝑒2𝑑+24𝑒4𝑑+21𝑒7𝑑)M'(t)=\dfrac{-1}{12}(2𝑒^{𝑑}+𝑒^{2𝑑}+6𝑒^{4𝑑}+3𝑒^{7𝑑})^{-2}(2𝑒^{𝑑}+2𝑒^{2𝑑}+24𝑒^{4𝑑}+21𝑒^{7𝑑})

=βˆ’21e7t+24e4t+2e2t+2et12(3e7t+6e4t+e2t+2et)2=-\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,

Mβ€²β€²(t)=eβˆ’t(441e12t+846e9t+9e7t+444e6t+72e4tβˆ’12e3t+4e2t+6et+4)12(3e6t+6e3t+et+2)3M''(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, Mβ€²(0)=βˆ’21e0+24e0+2e0+2e012(3e0+6e0+e0+2e0)2=βˆ’491728M'(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 Mβ€²β€²(0)=441+846+9+444+72βˆ’12+4+6+412(3+6+1+2)3=90710368M''(0)=\dfrac{441+846+9+444+72-12+4+6+4}{12(3+6+1+2)^3}=\dfrac{907}{10368}

Then, variance=Mβ€²β€²(0)βˆ’[Mβ€²(0)]2=90710368βˆ’(βˆ’491728)2=M''(0)-[M'(0)]^2=\dfrac{907}{10368}-(-\dfrac{49}{1728})^2

So, variance=0.08667=0.08667


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