Given the mgf $M_{X}(t)=\left(\frac{e^{t}+1}{2}\right)^4$, how can we determine the corresponding random variable $X$ (distribution + parameters)
I initially thought it kinda looked like the mgf of a Uniform distribution $\left(\frac{e^{tb}-e^{ta}}{t(b-a)}\right)$ but this doesn't really work and I don't know where else to go.
Recall that if $X$ is a discrete-valued random variable, then $$M_X(t) = \sum_{x\in\Omega} e^{tx} \Pr[X = x],$$ where $\Omega$ is the support of $X$. Therefore, $$M_X(t) = \left(\frac{e^t + 1}{2}\right)^4 = \frac{1}{16} e^{4t} + \frac{1}{4} e^{3t} + \frac{3}{8} e^{2t} + \frac{1}{4} e^t + \frac{1}{16}$$ is satisfied by an $X$ with PMF $$\Pr[X = 0] = \frac{1}{16} \\ \Pr[X = 1] = \frac{1}{4} \\ \Pr[X = 2] = \frac{3}{8} \\ \Pr[X = 3] = \frac{1}{4} \\ \Pr[X = 4] = \frac{1}{16}.$$ We can also write this as $$\Pr[X = x] = \binom{4}{x} \left(\frac{1}{2}\right)^x \left(1 - \frac{1}{2}\right)^{4-x}.$$ Doe this look familiar?