Function of Random Variable

Let $\mathbb{X}$ be a random variable, and $\mathbb{Y} = g(\mathbb{X})$ is a real valued function of $\mathbb{X}$.

example 1 If $\mathbb{X} \sim normal(\mu, \sigma^2)$, then $\mathbb{Y} = \frac{\mathbb{X}-\mu}{\sigma} \sim normal(0, 1)$.

proof 1 For the following proof, we appeal to c.d.f. Let $\mathbb{X} \sim F_X$ and $\mathbb{Y} \sim F_Y$, then $F_Y(y) = \mathbb{P}( \frac{\mathbb{X}-\mu}{\sigma} \leq y ) = \mathbb{P}( \mathbb{X} \leq \sigma y + \mu ) = F_X( \sigma y + \mu )$ $\Rightarrow f_Y(y) = F_Y'(y) = \frac{\mathrm{d}}{\mathrm{d}y}F_X( \sigma y + \mu ) = \sigma f_X( \sigma y + \mu ) = \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}$

proof 2 For the following proof, we appeal to m.g.f. Let $\mathbb{X} \sim F_X$ and $\mathbb{Y} \sim F_Y$, then $M_X(t) = \mathbb{E}[e^{tX}] = e^{\mu t + \frac{\sigma^2}{2}t^2}$ $M_Y(t) = \mathbb{E}[e^{t\frac{X-\mu}{\sigma}}] = e^\frac{-\mu t}{\sigma} \mathbb{E}[ e^\frac{tX}{\sigma}] = e^\frac{-\mu t}{\sigma}e^{\mu \frac{t}{\sigma} + \frac{\sigma^2}{2}(\frac{t}{\sigma})^2} = e^\frac{t^2}{2}$

And by the uniqueness property of moment generating function $\mathbb{Y} \sim normal(0, 1)$

example 2 If $\mathbb{X} \sim gamma(\alpha, \lambda)$, then $\mathbb{Y} = 2 \lambda \mathbb{X} \sim \chi(2 \alpha)$.

proof 1 Let $\mathbb{X} \sim gamma(\alpha, \lambda)$, $\mathbb{Y} = 2 \lambda \mathbb{X}$ $f_Y(y) = F_Y'(y) = \frac{\mathrm{d}}{\mathrm{d}y} \mathbb{P}( \mathbb{X} \leq \frac{y}{2 \lambda} ) = \frac{\mathrm{d}}{\mathrm{d}y} F_X( \frac{y}{2 \lambda} ) = \frac{1}{2 \lambda} f_X( \frac{y}{2 \lambda} )$ $= C (\frac{y}{2})^{\alpha-1} e^{-\frac{y}{2}} \sim \chi^2(2\alpha)$

proof 2 Let $\mathbb{X} \sim gamma(\alpha, \lambda)$, $\mathbb{Y} = 2 \lambda \mathbb{X}$ $M_X(t) = \mathbb{E}[e^{tX}] = (1-\frac{t}{\lambda} )^{-\alpha}$ $M_Y(t) = \mathbb{E}[e^{2 \lambda t \mathbb{X}}] = (1-2 t)^{-\alpha}$ so $Y \sim gamma(\frac{1}{2}, \frac{2\alpha}{2}) = \chi^2(2\alpha)$