Expectation of 4th moment under truncated condition First, some context. I'm currently working on some non-convex optimization algorithms. In particular, I've been following a few papers on Phase Retrieval using Wirtinger Flow and extending the results to non-gaussian sampling vectors. I'm getting nice results and now I want to prove that I can use newer versions of the algorithm (that provide better sampling accuracy and convergence properties) in my own setting.
In this particular follow-up paper, there's a step that is nagging me. This is a pure math step and there's no need to read the papers to compute this.
Let $\mathbf{a}_i\in\mathbb{R}^n$ be $K$ independent gaussian sampling vectors such that $\mathbf{a}_i\sim \mathcal{N}(\mathbf{0},\mathbf{I}_n)$, $\mathbf{x}\in\mathbb{R}^n$ a constant vector with $\|\mathbf{x}\|=1$,  and $\gamma$ a constant scalar, both independent of the sampling vectors.
This paper states that (Equations 150-151)
$$\mathbb{E}\Bigg\{\frac{1}{K}\sum_{i=1}^K\big|\mathbf{a}_i^T\mathbf{x}\big|^2\mathbf{a}_i\mathbf{a}_i^T\mathbf{1}_{\{|\mathbf{a}_i^T\mathbf{x}|\leq\gamma\}}\Bigg\}=\beta_1\mathbf{xx}^T+\beta_2\mathbf{I}_n$$
where
$$\beta_1=\mathbb{E}\{\xi^4\mathbf{1}_{\{|\xi|\leq\gamma\}}\}-\mathbb{E}\{\xi^2\mathbf{1}_{\{|\xi|\leq\gamma\}}\},\qquad \beta_2=\mathbb{E}\{\xi^2\mathbf{1}_{\{|\xi|\leq\gamma\}}\},$$
$\xi\sim\mathcal{N}(0,1)$, and $\mathbf{1}_{\{cond\}}$ is the indicator function, which is 1 if $cond$ is true and 0 otherwise.
For the life of me, I haven't been able to replicate the calculation, and I'm pretty sure that I'm missing an obvious fact here. Does anyone know how to compute this properly, step by step?
 A: I don't have a full argument, but maybe this can help. Ignoring the $K$ vector like we said in the comments, just think of one gaussian vector $(Y_1, \ldots, Y_n)$ where all $Y_i$ are independent standard normal. Also, denote $\xi = \sum_{k=1}^n Y_k x_k$. Notice that $x$ is a unit vector in $\mathbb{R}^n$ implies that $\xi$ is a standard normal variable.
The expected value above is a matrix $\Sigma$ with entries
$$ \Sigma_{ij} = \mathbb{E} [Y_i Y_j \xi^2 \mathbf{1}_{\{|\xi| \le \gamma\}}].$$
Expanding $\xi^2$, we can write that
$$
\Sigma_{i_0j_0}
= \mathbb{E} [Y_{i_0} Y_{j_0} \xi^2 \mathbf{1}_{\{|\xi| \le \gamma\}}]
= \sum_{i, j = 1}^n \mathbb{E} [Y_{i_0} Y_{j_0} Y_i Y_j \cdot 
  \mathbf{1}_{\{|\xi| \le \gamma\}}] x_i x_j.
$$
We just have to compute the coeficients. However, many of them should be zero.
Case $\gamma = \infty$. Notice that
$$
\mathbb{E} [Y_{i_0} Y_{j_0} Y_i Y_j]
$$
is zero if we have any index that appear as an odd power (remember that odd powers of a normal with mean zero also have mean zero). Then, if we have $i_0 \neq j_0$ we have that the only non-zero coeficients happen for $(i,j)$ being $(i_0, j_0)$ or $(j_0, i_0)$ and
$$
\Sigma_{i_0j_0}
= 2 \mathbb{E} [Y_{i_0}^2 Y_{j_0}^2] x_{i_0} x_{j_0}
= 2 x_{i_0} x_{j_0}.
$$
And for the diagonal terms we have that the non-zero coeficients occur when $i=j=i_0$ or when $i=j \neq i_0$, implying
$$
\Sigma_{i_0i_0}
= \mathbb{E} [Y_{i_0}^4 ] x_{i_0}^2 + 
  \sum_{i \neq i_0}\mathbb{E} [Y_{i_0}^2 Y_i^2] x_{i}^2
= 3 x_{i_0}^2  + \sum_{i \neq i_0} x_{i}^2
= 2 x_{i_0}^2 + 1,
$$
where in the last line we used that norm of $x$ is $1$.
