Given the m.g.f $M_x=\exp(3t+8t^2)$ of a continuous random variable $X$, find an m.g.f of the random variable $z=0.5(x-3)$, and use to find the mean and variance of $z$
My working:
We know that $\mu=(M_xt)', t=0$ upon doing this, I got $\mu=3$ and $E(x^2)=(M_xt)'', t=0$ and got $E(x^2)=25$.
And because $Var(x)=25-(3)^2=16$.
Am I on the right track? How do I find the mean and variance of $z$? Do I simply just put the values of that I obtained from above?
Recall that the MGF of a random variable $X$ is given by $$M_X(t) = \operatorname{E}[e^{tX}].$$ Thus, if $Z = \frac{1}{2}(X-3)$, then the MGF of $Z$ is given by $$M_Z(t) = \operatorname{E}[e^{tZ}] = \operatorname{E}[e^{t(X-3)/2}] = \operatorname{E}[e^{tX/2} e^{-3t/2}] = \operatorname{E}[e^{(t/2)X}] \operatorname{E}[e^{-3t/2}] = M_X(t/2) e^{-3t/2}.$$ Now substituting $$M_X(t) = e^{3t+8t^2},$$ being careful to first replace $t$ with $t/2$, we now have the MGF of $Z$. Then we calculate $$\operatorname{E}[Z] = M_Z'(0), \quad \operatorname{Var}[Z] = M_Z''(0)$$ using the function we just obtained, the calculation of which I have left to you as an exercise.