Moment generating function of quadruple-form of Gaussian RVs Let $X \sim N(0, I_d)$ be a $d$-dimensional Gaussian random variable. Let $\beta_1$ and $\beta_2$ be two $d$-dimensional vectors. I would like to compute expectation
\begin{align}
\mathbb{E}\Bigl \{  \exp\bigl[ -  (X^\top \beta_1)^2 \cdot (X ^\top \beta_2)^2 \bigr] \bigr\} .
\end{align}
I was wondering if there is a close-form solution. My conjecture is that this expectation is a function of the inner product of $\beta_1$ and $\beta_2$.
 A: I made the solution as complete as I could, but it is unfortunately incomplete. It can at least provide a start. 
Let 
$$\boldsymbol{\beta}_1=\begin{bmatrix}
b_{11} \\
b_{21} \\
\vdots \\
b_{d1}
\end{bmatrix}$$
and
$$\boldsymbol{\beta}_2=\begin{bmatrix}
b_{12} \\
b_{22} \\
\vdots \\
b_{d2}
\end{bmatrix}\text{.}$$
Also, let
$$\mathbf{X} = \begin{bmatrix}
X_1 \\
X_2 \\
\vdots \\
X_d\end{bmatrix}\text{.}$$
Notice that $\mathbf{X}$ has independent random variables. Then
$$\mathbf{X}^{\top}\boldsymbol{\beta}_1 = \sum_{i=1}^{d}X_ib_{i1}\sim \mathcal{N}\left(0,  \sum_{i=1}^{d}b_{i1}^{2}\right)$$
and similarly,
$$\mathbf{X}^{\top}\boldsymbol{\beta}_2 = \sum_{i=1}^{d}X_ib_{i2}\sim \mathcal{N}\left(0,  \sum_{i=1}^{d}b_{i2}^{2}\right)\text{.}$$
Given the above,
$$\dfrac{\sum_{i=1}^{d}X_ib_{i1}}{\sqrt{\sum_{i=1}^{d}b_{i1}^{2}}} \sim \mathcal{N}(0, 1) $$
and similarly,
$$\dfrac{\sum_{i=1}^{d}X_ib_{i2}}{\sqrt{\sum_{i=1}^{d}b_{i2}^{2}}} \sim \mathcal{N}(0, 1) $$
so
$$\left(\dfrac{\sum_{i=1}^{d}X_ib_{i1}}{\sqrt{\sum_{i=1}^{d}b_{i1}^{2}}}\right)^2\sim \chi^2_1$$
and
$$\left(\dfrac{\sum_{i=1}^{d}X_ib_{i2}}{\sqrt{\sum_{i=1}^{d}b_{i2}^{2}}}\right)^2\sim \chi^2_1$$
I'm not sure if anything else can be done beyond this, due to the dependence of these $\chi^2_1$ variables.
