Question

Given the moment-generating function Mx(t) = e^(3t+8t^2) , find the moment

generating function of the random variable Z = (X-3)/4, and use it to

determine the mean and the variance of Z.

Answer 1

$\begin{aligned} &\text { Solution: Let } x \text { has my } f \\ &\qquad \begin{aligned} M \times(t) &=e^{{3 t}+8 t^{2}} \\ \text { Let } z &=\frac{1}{2}(x-3) \\ M g f d f & z^{2} \\ M z(t) &=E\left(e^{z t}\right) \\ &=E\left(e^{\frac{1}{2}(x-3) t}\right) \\ &=E\left(e^{x \frac{t}{2}} e^{-3 \frac{t}{2}}\right) \\ &=e^{-3 t / 2} E\left(e^{x\left(\frac{t}{2}\right)}\right) \\ &=e^{-3 t / 2} M_{x}\left(\frac{t}{2}\right) \\ &=e^{-3 t / 2} \cdot e^{3\left(\frac{t}{2}\right)+8\left(\frac{t}{2}\right)^{2}} \\ &=e^{-3 t}+\frac{3 t}{2}+2 t^{2} \\ &=e^{0+2 t^{2}} \end{aligned} \\ &{\left[M_{z}(t)=e^{2 t^{2}}[\right.} \end{aligned}$

$\begin{aligned} E(z) &=\left.\frac{d}{d t} M z(t)\right|_{t=0} \\ &=\left.e^{2 t^{2}}(z t)\right|_{t=0} \\ &=e^{0}(4(0)) \\ E(z) &=0 \\ E\left(z^{2}\right) &=\left.\frac{d^{2}}{d t^{2}} P_{z}(t)\right|_{t=0} \\ &=\left.\frac{d}{d t} 4 t e^{2 t^{2}}\right|_{t=0}\{\operatorname{trc}(1)\\ &=\left[t(4 t) e^{2 t^{2}}+e^{2 t^{2}}(1)\right] \mid t=0 \\ &=4\left[0+e^{0}\right]=4(0+1) \\ E\left(z^{2}\right) &=4 \\ V(z) &=E\left(z^{2}\right)-[E(z)]^{2}=4-0 \\ V(z) &=4 \end{aligned}$