Answer to Question #243486 in Statistics and Probability for hardik

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

$$Var(X)=\sigma^2=E[X^2]-(E[X])^2$$

$$=M''(0)-(M'(0))^2$$

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

$$=\dfrac{2e^{-t}+1+6e^{2t}+3e^{5t}}{12}$$

$$M'(t)=\dfrac{-2e^{-t}+12e^{2t}+15e^{5t}}{12}$$

$$M''(t)=\dfrac{2e^{-t}+24e^{2t}+75e^{5t}}{12}$$

$$M'(0)=\dfrac{-2+12+15}{12}=\dfrac{25}{12}$$

$$M''(0)=\dfrac{2+24+75}{12}=\dfrac{101}{12}$$

$$Var(X)=\sigma^2=\dfrac{101}{12}-(\dfrac{25}{12})^2$$

$$=\dfrac{1212-625}{144}=\dfrac{587}{144}$$