MGF of X where $E(X^{2r})= (2r)!/2^rr!$ and $E(X^{2r-1}) = 0 $

Question

Let moment-generating function($mgf$) exists for given random variable $X$ and let $$E(X^{2r})= (2r)!/2^rr! \;\;\text{where}\;\;r=1,2,3,...$$ $$E(X^{2r-1}) = 0 \;\;\text{where}\;\;r=1,2,3, ...$$ then find $mgf$ of $X$

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I would like to derive pdf of $X$ from given above two equations. Is there any hint/advice for this problem?

Answer 1

$$M_{X}(t) = E(e^{tX}) = E\left(\sum\limits_{r=0}^{\infty} \frac{(tX)^{2r}}{(2r)!} + \sum\limits_{r=1}^{\infty} \frac{(tX)^{2r-1}}{(2r-1)!}\right) = E\left(\sum\limits_{r=0}^{\infty} \frac{(tX)^{2r}}{(2r)!} \right) = \sum\limits_{r=0}^{\infty}\frac{(t^{2}/2)^r}{r!}$$

$$\implies M_{X}(t) = e^{t^2/2}$$

Therefore, X follows normal distribution with mean $0$ and variance $1$. For $\mathcal{N}(\mu, \sigma^2)$, mgf is given by $e^{\mu t + \sigma^2t^2/2}$.

Note: I used the series expansion of $e^x$ in the first equation with a separation of terms with odd and even powers.