Let $\mathbf{X}$ be a p-variate random variable such that $M_\mathbf{X}(\mathbf{t})$ exists for all $\mathbf{t}\in\mathbb{R}^p$. Is the marginal moment generating function of the $k$-component of $\mathbf{X}$ simply $M_\mathbf{X}(\mathbf{t})$ evaluated at
$$\mathbf{t} = \left\{ \begin{array}00 \text{ if} & i \not= k \\ 1 \text{ if} & i = k \end{array} \right. $$ such that $i$ is the index of the components of $\mathbf{t}$?
It seems like that by definition, but i wonder if it's correct.
The context of this doubt is that I was searching for some properties of the multivariate normal distribution and one guy did the above to prove that the marginals are normal as well, as the MGF, density and characteristic function uniquely identify a distribution. The proof I know uses the Jacobian method instead, so that's why I'm asking.
The MGF of a random vector $X$ is given by $\Psi_X\left(\gamma\right) = \mathbb{E}\left(e^{\gamma^\top X}\right)$. A special case of $\gamma$ is a vector of zeros with the $k$th component equal to $t$. Plugging this into the definition yields $\mathbb{E}\left(e^{t X_k}\right)$, so yes, you can get back the univariate MGF in this way.