Question: If the sequence of moments of a random variable is given as $m_k = \frac{1}{2}(C^k+ (-1)^kC^k), c\in \mathcal{R}$, find the corresponding distribution.
The given answer is that the random variable X has distribution $P(X=C) = P(X=-C) = \frac{1}{2}$. I don't see why this is the case. For this particular random variable, the MGF can be computed as $$E(e^{kX}) = \frac{1}{2}*e^{Ck} + \frac{1}{2}*e^{-Ck} = \frac{1}{2}*(e^{Ck} + e^{-Ck})$$ I can't get from here to the answer. Any hints? Thank you.
Ok after giving it a thought I have it.
Let the moment generating function for the random variable X that has distribution $P(X=C) = P(X=-C) = \frac{1}{2}$ be $M_X(t) = E[e^{tX}] = \frac{1}{2}(e^{Ct} + e^{-Ct})$. Then the $k^{th}$ moment is given by $$m_k = \frac{\partial^{(k)}}{\partial t^{(k)}} M_x(t)|_{t=0} = \frac{1}{2}(C^k + (-1)^k C^k) $$
But I do not know how to go from the moments $m_k$ to the random variable $X$. Any other answers that can start form moments are appreciated.