Can a constant be a moment generating function of a random variable?

Question

I need to disprove that there exists a random variable $X$ such that its MGF is $M_X(t)=e$. Not sure how to approach this...

Answer 1

Assume that we have a random varible $\mathbf{X}$ and denote its probability distribution $f(x)$, for simplicity assume that this is a real random variable in one dimension,but any higher dimension euclidean space would also do the trick.

Now assume that the moment generating function is constant thus $M_X(t) = c$. From the definition of $M_X(t) = \mathbb{E}[e^{t\mathbf{X}}]$ we note that for $t= 0$ it holds :

$$M_X(0) = \mathbb{E}[e^{0 \cdot \mathbf{X}}] = \mathbb{E}[1] = 1$$

Now since we assumed that $M_X(t) $ is constant, it must be equal to one everywhere. Now we take the derivative of moment generating function, and if I understood the Leibniz integration rule correctly the derivative is:

$$M_X^{'}(t) = \int\limits_{-\infty}^{\infty} t e^{tx}f(x) \mathrm{d}x = t M_X(t)$$

But since the MGF ( moment-generating function ) is constant, this would have to be zero for all t, since this is not true, it can not be true that there exists a random variable whose MGF is constant

Answer 2

Hint

Note that $$ M_{X}(0)=Ee^{0\cdot X}=E(1)=1. $$