Consider the simple discrete distribution of the random variable \(X\) with probability function as given in equation (3.18). Make two transformations,

\[ Y=2 X+1 \]

and

\[ Z=X^{2}+1 \]

(a) Using the respective probability functions, compute \(\mathrm{E}(X)\), \(\mathrm{V}(X), \mathrm{E}(Y), \mathrm{V}(Y), \mathrm{E}(Z)\), and \(\mathrm{V}(Z)\).

(b) Now, using \(\mathrm{E}(X)\) and \(\mathrm{V}(X)\), and using the transformations only, compute \(\mathrm{E}(Y), \mathrm{V}(Y), \mathrm{E}(Z)\), and \(\mathrm{V}(Z)\).

\(f_X(x)= \begin{cases}1 / 4 & \text { for } x=-1 \\ 1 / 2 & \text { for } x=0 \\ 1 / 4 & \text { for } x=1 \\ 0 & \text { otherwise }\end{cases} \tag{3.18}\)