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I am trying to find the MGF of a random vector where $Z_i\overset{iid}\sim N(0,1)$

the new Random Vector is $X\overset{d}=\mu+AZ$ where $\mu\in\mathbb{R^k}$ and $A\in\mathbb{R}^{k\times m}$

where $A$ is not assumed to be full rank. I know that this will be a normal but I was considering what the variance of this random variable would be. I know that if $A$ is orthonormal then independence follows where $A$ is assumed to be a square matrix. If the Matrix $A$ is not a square matrix then if I assume $\text{rank}(A)=m$ then I know that the $\text{rank}(A^T)=m$ and the variance of $X$ is $AA^T$. If I consider whether $AA^T$ is positive definite that is $y^{T}AA^Ty=0$ then $y=0$. I can show that $A^Ty=0$ by using the euclidean norm but since $\text{rank}(A^T)=m$ then $y$ does not have to be $0$ if $k>m$. I can't see a scenario where this is not a degenerate random vector when $A$ is not square of full rank. Am I missing something? How do I go about solving for the MGF in general case without imposing assumptions?

StubbornAtom
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nvm
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2 Answers2

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$X$ is multivariate normal with mean $\mu$ and covariance matrix $AA^\top$, regardless of the shape and rank of $A$. You can then immediately write down the MGF: $M_X(t) = e^{\mu^\top t + \frac{1}{2} t^\top AA^\top t}$.

You seem to care more about whether $X$ is degenerate. The following are equivalent.

  1. $X$ is not degenerate
  2. $AA^\top \in \mathbb{R}^{k \times k}$ has rank $k$
  3. $A \in \mathbb{R}^{k \times m}$ has rank $k$ (see here)

Now, note that the third statement here implies $k \le m$. Phrased in another way, if $m < k$ then $\text{rank}(A) \le m < k$ and thus $X$ will be degenerate.

angryavian
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Recall that definition of the MGF in general is $M_X(t)=Ee^{(X,t)}$. This makes no assumptions about the covariance structure of $X$, so you can just plug it in. In this case you would get

\begin{eqnarray*} M_{AZ+\mu}(t)&=&Ee^{(AZ+\mu,t)}\\ &=&Ee^{(AZ,t)+(\mu,t)}\\ &=&e^{(\mu,t)}Ee^{(Z,A^Tt)}\\ & = & e^{(\mu,t)}M_{Z}(A^Tt) \end{eqnarray*} and i assume you know how to find $M_Z(t)$

Simon Segert
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