$X$ has MGF $m_{x}(t)=e^{2t}$ for any $t in mathbb{R}$. What is the distribution of $X$?

My instructor went about it like this:

$m_{x}(t)=E(e^{xt})=Ee^{2t}+e^{2t}$ Here I am really confused, how did $E(e^{xt})$ become what's on the right hand side of the equation? I wonder if I copied it down wrong…but can someone please explain what he was trying to get at here?

So $P(X=2)=1$ Also, this is assuming that $X$ is a continuous variable, correct? How can I tell from the problem that $X$ is continuous, not discrete?

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#### Best Answer

$m_{x}(t)=E(e^{xt})=Ee^{2t}+e^{2t}$ Here I am really confused, how did $E(e^{xt})$ become what's on the right hand side of the equation?

Simply, the expectation of a constant $c$ is $c$.

Since $mathbb E[e^{tX}]=e^{2t}$, by multiplying by the constant $e^{-2t}$ on both sides we obtain $mathbb E[e^{t(X-2)}]=1$ for each $t$. Differentiating two times and taking the value $t=0$, we obtain that $mathbb E[(X-2)^2]=0$, hence $mathbb P(X=2)=1$.

We didn't make the assumption that $X$ is continuous.