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I have to solve the following question:

Let $X$ follow the distribution with moment generating function $M_X(t)$ and Let $Y = aX + b$ follow the distribution with moment generating function $M_Y(t)$.

Show that $$M_Y(t) = e^{bt}M_X(at).$$

Now, I know that I need to find the density $f_Y(y)$ of $Y$ in terms of $f_X$ which is
$$f_Y = \frac{1}{a}f_X(\frac{y-b}{a}).$$
The question is, how do I identify the pdf of $f_X$ if it has not been stated in the problem what kind of distribution $X$ has?
Is there a generic pdf that can be used when the distribution of a random variable is not stated?

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You don't need to use the density to show that $M_Y(t) = e^{tb} M_X(at)$ (if it is your question).
By definition, the moment generating function of a random variable $X$ is: $$ M_X(t) = \mathbb{E}[ e^{tX} ] $$

Since $Y= aX + b$ we have \begin{align*} M_Y(t) &= \mathbb{E} [ e^{tY} ] \\ &=\mathbb{E} [ e^{t(aX+b)} ] \\ &= \mathbb{E} [ e^{taX + tb} ] \\ &= \mathbb{E} [ e^{taX} e^{tb}] \\ &= e^{tb}\mathbb{E} [ e^{taX} ] \ \ (\text{since} \ e^{tb} \ \text{is constant}) \\ &= e^{tb} M_X(ta) \end{align*}

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Here is my suggestion: $M_Y(t)=\mathbb{E}[e^{tY}] = \mathbb{E}[e^{t(aX+b)}]=\mathbb{E}[e^{atX}e^b]=e^{tb}\mathbb{E}[e^{atX}]=e^{tb}M_X(at)$

If you wish, you can rewrite the mathematical expectations in terms of integrals over the pdf.

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