Give the mgf of a random variable having pdf $$f_X(x) =e^{−2|x|}I_{(−\infty,\infty)}(x)$$
I found a very similar problem here but I wanted to make sure I'm applying it correctly:
$$ \begin{align} M_X(t) &=\operatorname E(e^{tX}) \\\\ & =\int_{-\infty}^0 f(x) e^{tx}\ dx + \int_0^\infty f(x) e^{tx}\ dx\\\\ & =\int_{-\infty}^0 e^{2x} e^{tx}\ dx + \int_0^\infty e^{-2x} e^{tx}\ dx\\\\ &=\frac{1}{t+2}+\frac{1}{2-t} \end{align} $$
Following the pattern in the problem I referenced, would the support of $t$ be from $(-2,2)$? If so, what is the reasoning for that?
I wouldn't use the word support except when talking about the closure of the subset of the domain on which the function is not zero, or a related concept also yielding a closed set.
However, the domain of $M_X(t) = \operatorname E(e^{tX})$ is the set of values of $t$ for which $\operatorname E(|e^{tX}|)<+\infty,$ and since $t$ and $X$ are real, and the function $e^s$ is positive for all $s\in\mathbb R,$ that's the same as the set of values of $t$ for which $\operatorname E(e^{tX})<+\infty.$ If $t\ge2$ then $$ \int e^{-2s} e^{tx} \,dx = +\infty $$ and similarly the integral over the negative half-line is $+\infty$ if $t\le-2.$ So the domain of $M_X$ is the open interval $(-2,2).$