Left tail of convex combinations of $\chi_1^2$

Question

Suppose $a_1,...,a_n\geq0, \sum_{i=1}^na_i=1$ and $Z_1,...,Z_n$ are i.i.d. standard normal, what is a sharp upper bound of the following probability as $\delta\to0$ and what is the order? $$\mathbb{P}(\sum_{i=1}^na_iZ_i^2\leq\delta)$$ How will the distribution of $a_1,...,a_n$ affect the upper bound?

Answer 1

$\renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta}$

One can use formula (1.14) in Rozovsky: \begin{equation} \P(\sum_1^n X_i\le r)\sim k_n\prod_1^n \P(X_i\le r) \end{equation} for $r\downarrow0$, where the $X_i$'s are independent positive random variables with \begin{equation} \P(X_i\le x)=\ell_i(x)x^{b_i} \end{equation} for $x>0$, the $\ell_i$'s are functions slowly varying at $0$, $b_i>0$, and $k_n:=\prod_1^n\Ga(1+b_i)/\Ga(1+\sum_1^n b_i)$.

In our case, $X_i=a_i Z_i^2$, $\P(X_i\le x)\sim\sqrt{\frac{2x}{\pi a_i}}$ for $x\downarrow0$, and $b_i=1/2$, whence \begin{equation} \P(\sum_1^n a_i Z_i^2\le\de)\sim\frac{(\de/2)^{n/2}}{\Ga(1+n/2)\,\prod_1^n\sqrt{a_i}} \end{equation} as $\de\downarrow0$.

Answer 2

Let $X = \sum_{i=1}^n a_i Z_i^2$. If $m = \min(a_1,\ldots,a_n)$ and $M = \max(a_1,\ldots,a_n)$, we have $m A \le X \le M A$ where $A$ has $\chi^2$ distribution with $n$ degrees of freedom. Thus $$\mathbb P(A \le \delta/M) \le \mathbb P(X \le \delta) \le \mathbb P(A \le \delta/m)$$ and so $\mathbb \delta^{n/2} P(X \le \delta)$ is bounded above and below as $\delta \to 0+$.