Answer
(a) Normal, $\mu=0$, $\sigma^2=12$
(b) Exponential, $\lambda=2$
(c) Binomial, $n=4$, $p=\frac{1}{2}$
(d) Geometric, $p=0.3$
Work Step by Step
(a)
$\begin{align*}
M_Y(t) &= e^{6t^2} \\
&= e^{0t + \frac{1}{2}\cdot 12t^2} \\
M_Y(t) &= e^{\mu t + \frac{1}{2}\cdot\sigma^2t^2},
\end{align*}$
where $\mu = 0$ and $\sigma^2=12$.
The mgf of $Y$ is that of a normal random variable with mean $\mu=0$ and variance $\sigma^2=12$ (see Example 3.12.4, p. 208).
Thus, $Y\sim N(\mu=0,\sigma^2=12)$ so that the pdf of $Y$ is Normal with mean $\mu=0$ and variance $\sigma^2=12$.
(b)
$\begin{align*}
M_Y(t) &= 2/(2-t) \\
M_Y(t) &= \frac{\lambda}{\lambda-t},
\end{align*}$
where $\lambda =2$.
The mgf of $Y$ is that of an exponential random variable with parameter $\lambda = 2$ (see Example 3.12.3, p. 208).
Thus, $Y\sim \text{Exponential}(\lambda = 2)$ so that the pdf of $Y$ is Exponential with parameter $\lambda=2$.
(c)
$\begin{align*}
M_Y(t) &= \left(\frac{1}{2} + \frac{1}{2}e^t\right)^4 \\
&= \left(\left(1-\frac{1}{2}\right) + \frac{1}{2}e^t\right)^4 \\
M_Y(t) &= \left((1-p) + pe^t\right)^n,
\end{align*}$
where $n=4$ and $p=\frac{1}{2}$.
The mgf of $Y$ is that of a binomial random variable with parameters: number of trials $n=4$ and probability of success $p=\frac{1}{2}$ (see Example 3.12.2, p. 207).
Thus, $Y\sim \text{Binomial}(n=4,p=\frac{1}{2})$ so that the pdf of $Y$ is a Binomial distribution with parameters $n=4$ and $p=\frac{1}{2}$.
(d)
$\begin{align*}
M_Y(t) &= \dfrac{0.3e^t}{1-0.7e^t} \\
&= \dfrac{0.3e^t}{1-(1-0.3)e^t} \\
M_Y(t) &= \dfrac{pe^t}{1- (1-p)e^t} \\
\end{align*}$
where $p=0.3$.
The mgf of $Y$ is that of a geometric random variable wherein each trial has a probability of success $p=0.3$ (see Example 3.12.1, p. 207).
Thus, $Y\sim \text{Geometric}(p=0.3)$ so that the pdf of $Y$ is Geometric with a probability of success $p=0.3$.
