Given that the mean vector of the random vector $\mathbf{X}$ is $m_{X}$ and the covariance matrix is $Λ_{x}$, we define the random variable $Y$ as
$$
Y=G^{T} X
$$
If for any non-zero real vector $G$ such that $E(Y^{2}) < \infty$, $Y$ is Gaussian, try to find the joint characteristic function $\phi(v)$ of $\mathbf{x}$ is ().

Options:
A. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )\neq \phi_{Y} ( \boldsymbol{v} ) \left| \right._{\boldsymbol{v} G^{\mathsf{T}}=\boldsymbol{v}^{\mathsf{T}}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{T} \boldsymbol{m}_{X}+\frac{1} {2} \boldsymbol{v}^{T} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
B. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )=\phi_{Y} ( v ) \left| \right._{\boldsymbol{v}G^{\mathsf{T}}=v^{\mathsf{T}}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{T} \boldsymbol{m}_{X}-\frac{3} {2} \boldsymbol{v}^{T} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
C. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )= \phi_{Y} ( \boldsymbol{v} ) |_{\boldsymbol{v} G^{\mathsf{T}}=\boldsymbol{v}^{T}}}}} \\ {{{=\operatorname{e x p} \left\{i n \boldsymbol{v}^{T} \boldsymbol{m}_{X}-\frac{1} {4} \boldsymbol{v}^{T} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
D. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )=\phi_{Y} ( \boldsymbol{v} ) \left| \right._{\boldsymbol{v} G^{\mathsf{T}}=\boldsymbol{v}^{\mathsf{T}}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{m}_{Y}-\frac{3} {2} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
E. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )=\phi_{Y} ( \boldsymbol{v} ) \left| \right._{\boldsymbol{v}^{\mathsf{T}} G=\boldsymbol{v}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{m}_{X}-\frac{1} {2} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{\Delta}_{Y} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
F. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )=\phi_{Y} ( \boldsymbol{v} ) \left| \right._{\boldsymbol{v}G=\boldsymbol{v}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{m}_{X}+\frac{1} {2} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{\Delta}_{Y} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
G. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )=\phi_{Y} ( v) \left| \right._{\boldsymbol{v} G^{\mathsf{T}}=\boldsymbol{v}^{\mathsf{T}}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{m}_{X} -\frac{1} {3} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
H. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )=\phi_{Y} ( v ) \left| \right._{\boldsymbol{v} G^{T}=v^{\mathsf{T}}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} v^{\mathsf{T}} \boldsymbol{m}_{X}-\frac{2} {3} \boldsymbol{v}^{T} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
I. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} ) \neq \phi_{Y} ( \boldsymbol{v} ) \left| \right._{\boldsymbol{v} G^{\mathsf{T}}=\boldsymbol{v}^{\mathsf{T}}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{m}_{Y}+\frac{1} {2} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
J. $$
\begin{array} {c} {{{\phi_{X} ( \boldsymbol{v} )=\phi_{Y} ( \boldsymbol{v} ) \left| \right._{\boldsymbol{v} G^{\mathsf{T}}=\boldsymbol{v}^{\mathsf{T}}}}}} \\ {{{=\operatorname{e x p} \left\{\operatorname{i n} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{m}_{X}-\frac{1} {2} \boldsymbol{v}^{\mathsf{T}} \boldsymbol{\Delta}_{X} \boldsymbol{v} \right\}}}} \\ \end{array}
$$
Answer: J