Normal distribution in equality

Question

Let $p(x)=a_1x_1+a_2x_2+. . . .a_n x_n$ be a polynomial such that $\sum_ia_i^2=1,$ each $x_i \sim N(0,1)$. then we know that $p(x) \sim N(0,1)$. How can we bound $\Pr_{x\in N^n}\left(\left|p(x)-t\right|\leq \epsilon\right)$, for any $t \in \mathbb{R}$

Answer 1

Assuming the $x_i$ are independent. $\Pr_{x\in N^n}(|p(x)-t| \le \epsilon)= \Pr_{x\in N^n}(t-\epsilon\le a_1x_1+\ldots+a_nx_n \le t+\epsilon)=$

$$\int_{\epsilon\le a_1x_1+\ldots+a_nx_n \le t+\epsilon}f(x_1,\ldots,x_n) dx_1\ldots dx_n$$ where $f$ is the probability distribution. Pick some orthogonal basis $u_1\ldots u_n$ where $u_1 =(a_1,\ldots, a_n)$ and do change of variables.

$$\int_{\epsilon\le a_1x_1+\ldots+a_nx_n \le t+\epsilon}f(x_1,\ldots,x_n) dx_1\ldots dx_n=$$ $$ \int_{t-\epsilon\le u_1 \le t+\epsilon}\left(\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}f(u_1,\ldots,u_n) du_2\ldots du_n\right)du_1= $$ $$ \int_{t-\epsilon\le u_1 \le t+\epsilon} g(u_1)du_1 $$ where $g(u_1)$ is the result of computing the inner integrals. Finally upper bound that using the maximum of $g(u_1)$ over $t-\epsilon\le u_1 \le t+\epsilon$

$$ \int_{t-\epsilon\le u_1 \le t+\epsilon} g(u_1)du_1 \le 2\epsilon\left( \max_{t-\epsilon\le u_1 \le t+\epsilon}g(u_1)\right)$$