Let $X$ be a random variable such that $M_X(t)=e^t M_X(-t)$. Find $E(X)$ and $E\left(X^2\right)$.
I know the general procedure that to find the expected value, the first order derivative needs to be taken and then t must be set to 0.
$M_X(t)^{\prime}=e^t M_X(-t)-e^t M_X^{\prime}(-t)$
The solution is $E[X] = \frac{1}{2}$, but I don't see how I can evaluate this question without an expression for the MGF or a distribution for the random variable.
Question: Is there sufficent information to answer this question?
Question 2:
\begin{aligned} &M_x^{(2)}(t)=e^t\left(M_x^{(2)}(-t)-2 M_x^{(1}(-t)+M x(-t)\right) \\ &M_x^{(2)}(0)=M_x^{(2)}(0)-2 M_x^{(1)}(-t)+M x(0) \end{aligned}
If I differentiate two times, It seems like there is no expression for the second derivative. Does this mean that $E[X]$ is undefined?
The hypothesis says that $X$ and $_X$ have the same MGF. So they have the same distribution. In particular, $E=E(1-X)$ which gives $EX=\frac 1 2$.
There is not enough information to find the second moment, as you have suspected. Here are two examples where the hyptohesis is satisfied but there are different values for $EX^{2}$:
Ex. 1) $X$ Bernoulli. In this case $EX^{2}=\frac 12$.
Ex. 2) $X \sim U(0,1)$. Here $EX^{2}=\frac 1 3$.