Transformation of the MGF

Question

Let X be a continuous random variable with the probabilty density function given by

$f_X(x) = \frac{24}{x^4} , x>2$.

Let $Y = \frac{1}{X^3}$. Compute the mgf and pdf of Y.

How do I find the mgf of Y? I know that I can find the pdf of Y using the formula $$f_Y(y)=f_X(g^{-1}(y)) \vert\frac{\partial}{\partial y}g^{-1}(y)\vert.$$

Update: I got the pdf of Y as Y~Uniform$[0, \frac{1}{8}]$

Answer 1

Once you have the density of $Y$, simply compute $$M_Y(t) = \operatorname{E}[e^{tY}] = \int_{y=0}^{1/8} 8e^{ty} \, dy.$$

Answer 2

Correct, more or less! $Y\sim U\left(0;\frac{1}{8}\right)$

(zero and 8 have to be excluded from the support)

Now the MGF is known but if you want to calculate it just use the definition

$$MGF_Y=\mathbb{E}\left(e^{tY}\right)$$