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Whilst doing exercises on probability, I stumbled upon the following one:

Suppose $X$ is a random variable and its moment generating function $M_x(t)$ is a polynomial, then the expectation $\mathbb{E}\left(X^{2k}\right) =0$ for a certain $k$ in $\mathbb{N}$.

Now, I do now that $\mathbb{E}\left(X^{2k}\right)$ is defined as: $\mathbb{E}\left(X^{2k}\right)=M_x(t)^{(2k)}(0)$. Furthermore, I do understand that if the polynomial is of odd degree, it is easy to prove.

How can I proceed?

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    The degree of the polynomial doesn't matter. How many derivatives of a polynomial are nonzero? Finitely or infinitely many? – 2017-01-20
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    Infinitely many are zero of course, I did think of that of course. Using that, it directly follows that there always is such a $k$ (regardless of being odd or even), yet I found this a silly proof. – 2017-01-20
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    A funny exercise. It is obvious that $X=0$ almost surely (hence $M=0$) under this assumption. – 2017-01-20
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    Could you extend on why that is obvious? – 2017-01-20
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    If a non-negative random variable ($X^{2k}$ in this case) has zero mean, then it is zero almost surely. – 2017-01-20
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    Thank you! I did not take the even power in mind, which results in the random variable $X^{2k}$ only taking positive values, so that $\mathbb{P}(X^{2k}=0)=1$ , which is only the case if and only if , $\mathbb{P}(X=0)=1$, right? – 2017-01-20
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    This is exactly what I'm saying. – 2017-01-20
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    @zhoraster nice observation. – 2017-01-21

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