## Probability

### The Analysis of Data, volume 1

Random Vectors: Moment Generating Function

## 4.8. Moment Generating Function

Proposition 4.8.1. If $X_1,\ldots,X_n$ are independent RVs with mgfs $m_{X_i},\ldots,m_{X_n}$, then the mgf of their sum is $m_{\sum_{i=1}^n X_i} (t) = \prod_{i=1}^n m_{X_i}(t)$
Proof. \begin{align*} m_{\sum_{i=1}^n X_i} (t) &= \E\left(\exp\left( t\sum_{i=1}^n X_i\right)\right) \\ &= \E\left(\prod_{i=1}^n \exp(tX_i)\right) \\ &=\prod_{i=1}^n \E(\exp(tX_i))\\ &=\prod_{i=1}^n m_{X_i}(t). \end{align*}
Definition 4.8.1. We define the moment generating function of a random vector $\bb{X}$ as follows $m_{\bb{X}}:\mathbb{R}^n\to \mathbb{R}, \qquad m_{\bb{X}}(\bb{t}) = \E(\exp(\bb{t}^{\top}\bb{X})).$

As in the one dimensional case the mgf uniquely characterizes the distribution of the random vector.

Proposition 4.8.2. Suppose that $\bb{X}$, $\bb{Y}$ are two random vectors whose mgfs $m_{\bb{X}}(\bb{t})$, $m_{\bb{Y}}(\bb{t})$ exist for all $\bb{t}\in B_{\epsilon}(\bb 0)$ for some $\epsilon>0$. If $m_{\bb{X}}(\bb{t})=m_{\bb{Y}}(\bb{t})$ for all $\bb{t}\in B_{\epsilon}(\bb 0)$, then the cdfs of $\bb X$ and $\bb Y$ are identical (and consequentially also the pdf or pmf functions are identical).
Proof. The proof is similar to the proof of the one dimensional case (Proposition 2.4.2).