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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?

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2 Answers 2

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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 $E(X^2) \ge (E X)^2=1/4.$

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    $\begingroup$ Thank you very much! I cannot overstate the relevance of this platform that enables beginners to interact with experts $\endgroup$
    – Tim
    Sep 23, 2022 at 11:49
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You are right in concluding that the second derivitive term is lost. Therefore, it is only possible to give a rather general answer.

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