The question simply says:
Derive the MGF with a single variable (don't give memorized equation).
I'm not entirely sure what this question is asking, and I know how to derive the MGF if I knew that the random variable was binomial or geometric for example. However, I don't know how to derive it for a single variable. What I'm thinking is this:
$$ \begin{split} e^x &= \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!} + \ldots \\ e^{xt} &= 1+xt+\frac{x^2t^2}{2!}+\frac{x^3t^3}{3!} + \ldots \\ M(t) = E(e^{xt}) &= 1+tE(x)+\frac{t^2}{2!}E(x^2)+\frac{t^3}{3!}E(x^3) + \ldots \end{split} $$
I'm not sure if this is the answer?
Is it asking to derive the general formula for the MGF for a discrete random variable (since we haven't done continuous r.v.'s in class yet).
I know that for a discrete pmf, the MGF is:
$$M(t)=\sum_{i=0}^{\infty}e^{tx_i}p_i$$
EDIT:
So should I say then that:
$$M_{X}(t)=E(e^{tX})$$
$$=E\Bigg(\sum_{r=0}^{\infty} \frac{(tx)^r}{r!}\Bigg)$$
$$=\sum_{r=0}^{\infty} \frac{E(X^r)}{r!}*t^r$$