Answer to Question #213998 in Statistics and Probability for max

Question

The moment generating function of a random variable Y is given by M(t)=(5e^-t-4)^-t. Find the standard deviation of y

Answer 1

$M(t)=(5e^{-t}-4)^{-t}$

$M^\prime(t)=\dfrac{\frac{5t\mathrm{e}^{-t}}{5\mathrm{e}^{-t}-4}-\ln\left(5\mathrm{e}^{-t}-4\right)}{\left(5\mathrm{e}^{-t}-4\right)^t}$

$E(x)=M^\prime(0)=0$

$M^{\prime\prime}(t)=\dfrac{\left(16\mathrm{e}^{2t}-40\mathrm{e}^t+25\right)\ln^2\left(5\mathrm{e}^{-t}-4\right)+\left(40t\mathrm{e}^t-50t\right)\ln\left(5\mathrm{e}^{-t}-4\right)+\left(20t-40\right)\mathrm{e}^t+25t^2+50}{\left(5\mathrm{e}^{-t}-4\right)^t\left(4\mathrm{e}^t-5\right)^2}$

$E(x^2)=M^{\prime\prime}(0)=10$

$\sigma=\sqrt{E(x^2-(E(x))^2}$

$=\sqrt{10-0}$

$=\sqrt{10}$