Discussion
If \(P_{X}(t)\) is the probability generating function (pgf) of random variable X, then pgf of Y = 3X + 2 is :
Answer: A
The pgf of Y is \(t^{2} P_{X}(t^{3})\).
By definition \(P_{Y}(t) = E(t^{Y})\), so
\[ \begin{aligned} P_{Y}(t) &= E\left(t^{3X + 2}\right) = t^{2}\,E\left((t^{3})^{X}\right) \\ &= t^{2} P_{X}(t^{3}) \end{aligned} \]
The constant shift 2 becomes the factor \(t^{2}\) and the scale 3 becomes the argument \(t^{3}\).
\(P_{X}(t^{3})\) is the pgf of 3X, without the shift. \(t^{3}P_{X}(t^{2})\) belongs to \(2X + 3\), with the roles swapped, and \(2tP_{X}(t)\) is not a pgf at all, since it equals 2 at \(t = 1\).
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