# How do I find the second order moment with a MGF?

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 $$n$$th moment, the $$n$$th 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 $$E[X] = \frac{1}{2}$$

Question 1:

For the second order moment:

\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)}(0)+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 $$E[X^2]$$ is undefined?

Note that any variable with a finite MGF in a neighborhood of $$0$$ must have finite moments of all orders.
Let $$Y=X-1/2$$. Then the hypothesis is equivalent to $$M_Y(t)=M_Y(-t)$$ for all $$t$$, which in turn is equivalent to $$Y$$ and $$-Y$$ having the same law. Thus $$E(Y)=-E(Y)=0$$, but there is no information on the variance of $$Y$$ (which equals the variance of $$X$$) except that it is finite.
For example, $$Y$$ could have $$N(0,\sigma^2)$$ distribution for any $$\sigma \ge 0$$. Thus all one can say about $$E(X^2)$$ is that $$E(X^2) \ge (E X)^2=1/4.$$