Answer

Since the two random variables are independent

\[ \begin{aligned} E[e^{tY}] &= E[e^{t(2X_{1} + 6X_{2})}] = E[e^{(2t)X_{1}}]E[e^{(6t)X_{2}}]\newline &= \phi_{1}(2t)\phi_{2}(6t) \end{aligned} \]

For a \(Gamma(\alpha, \lambda)\), the mgf is

\[ \begin{aligned} mgf = \roundbr{\frac{\lambda}{\lambda - t}}^{\alpha} \end{aligned} \]

Hence,

\[ \begin{aligned} E[e^{tY}] &= \roundbr{\frac{1/3}{1/3 - (2t)}}^{3}\roundbr{\frac{1}{1 - (6t)}}^{5}\newline &= \roundbr{\frac{1}{1 - 6t}}^{8} = \roundbr{\frac{1/6}{1/6 - t}}^{8} \end{aligned} \]

which is the mgf of \(Gamma(8, 1/6)\)