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?



Left tail of convex combinations of $\chi_1^2$
$\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\Ga}{\Gamma}
\newcommand{\de}{\delta}$


One can use formula (1.14) in [Rozovsky](https://link.springer.com/article/10.1007/s10958-006-0251-2): 
\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 $\de\downarrow0$. 


