Moment generating function of quadruple-form of Gaussian RVs

Question

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 $\beta_2$.

Answer 1

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.