Asymptotic scaling of mean and variance for non-central chi distribution

Question

Define $Y \equiv \sqrt{\sum_{i=1}^k(\frac{ X_i}{\sigma_i})^2}$, with $X_i \sim \mathcal{N}(\mu_i, \sigma_i^2)$ and independents.

It is known that $Y$ is distributed as a non-central chi (Noncentral chi distribution). Let $\lambda = \sqrt{\sum_{i=1}^k \left( \frac{\mu_i}{\sigma_i} \right)^2}.$ Now assume that $$\lim_{k \rightarrow \infty} \frac \lambda {\sqrt k} = L,$$ with $L$ being a finite constant, i.e. $0<L<\infty$.

Question

How do the mean and variance of the distribution scale with $k$ in the limit of $k \rightarrow \infty$?

Answer 1

$\newcommand{\la}{\lambda}$Up to the equality in distribution, \begin{equation*} Y=\sqrt{\sum_1^k(Z_i+L_k)^2}, \end{equation*} where $Z,Z_1,\dots,Z_k$ are independent standard normal random variables (r.v.'s) and \begin{equation*} L_k:=\la/\sqrt k\to L \end{equation*} (as $k\to\infty$).

By the law of large numbers, \begin{equation*} \frac{Y^2}k=\frac1k\sum_1^k(Z_i+L)^2+(L_k-L)^2+2(L_k-L)\frac1k\sum_1^k (Z_i+L) \\ \to E(Z+L)^2=1+L^2 \end{equation*} and hence \begin{equation*} \frac{Y}{\sqrt k}\to\sqrt{1+L^2} \end{equation*} in probability and hence in distribution.

Also, $Var((Z+L_k)^2)=2+4L_k^2$ and hence $Var(Y^2)=k\,Var((Z+L_k)^2)=2k+4\la^2\sim2k(1+2L^2)$. So, by the central limit theorem (or, more precisely, by, say, the Berry--Esseen inequality), \begin{equation*} \frac{Y^2-(k+\la^2)}{\sqrt{2k(1+2L^2)}}\to Z\sim N(0,1) \end{equation*} in distribution. So, by Slutsky's theorem, \begin{equation*} Y-\sqrt{k+\la^2}=R_k:=\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(Y+\sqrt{k+\la^2})/\sqrt{2k}} \to\frac{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z \end{equation*} in distribution.

Also, by (say) the Rosenthal inequality, \begin{equation*} ER_k^4\le E\Big(\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(\sqrt{k+\la^2})/\sqrt{2k}}\Big)^4=O(1). \end{equation*} So, by uniform integrability, \begin{equation*} E(Y-\sqrt{k+\la^2})\to E\frac{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z=0 \end{equation*} and \begin{equation*} E(Y-\sqrt{k+\la^2})^2\to E\Big(\frac{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z\Big)^2=\frac{{1+2L^2}}{2(1+L^2)}. \end{equation*} Thus, \begin{equation*} EY=\sqrt{k+\la^2}+o(1) \tag{1}\label{1} \end{equation*} and \begin{equation*} Var\, Y= E(Y-\sqrt{k+\la^2})^2-(E(Y-\sqrt{k+\la^2}))^2\to\frac{{1+2L^2}}{2(1+L^2)}. \tag{2}\label{2} \end{equation*}

Answer 2

For simplicity, I take equal $\mu_i=\mu$, $\sigma_i^2=\sigma$. The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward, $k(\mu/\sigma)^2\mapsto\sum_{i=1}^k(\mu_i/\sigma_i)^2={\cal O}(k).$

For large $k$ the variable $Y^2$ has a Gaussian distribution with mean $$M=k[(\mu/\sigma)^2+1]$$ and variance $$V=k\bigl[(\mu/\sigma)^4+6(\mu/\sigma)^2+3-(M/k)^2\bigr]=4k[(\mu/\sigma)^2+1/2].$$ It follows that, to leading order in $1/k$, $$\mathbb{E}[Y] =\int_{-\infty}^\infty dz\, (2\pi V)^{-1/2}e^{-z^2/2V}(\sqrt{M}+\tfrac{1}{2}zM^{-1/2}-\tfrac{1}{8}z^2M^{-3/2})=M^{1/2}-\tfrac{1}{8}VM^{-3/2}$$ $$=k^{1/2}\sqrt{(\mu/\sigma)^2+1}+{\cal O}(k^{-1/2}),$$ $$\text{var}\,[Y]=M-\mathbb{E}[Y]^2=\frac{V \left(16 M^2-V\right)}{64 M^3}$$ $$=\frac{(\mu/\sigma)^2+1/2}{(\mu/\sigma)^2+1}+{\cal O}(k^{-1}).$$ So the mean of $Y$ scales as $\sqrt{k}$ while the variance tends to a constant in the limit $k\rightarrow\infty$.

As a check, I plot the mean and variance of $Y$ as a function of $k$ for $\mu=1=\sigma^2$: exact = gold, asymptotic = blue:

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The question has been edited with a request for the case that the $X_i$'s are non-Gaussian. The distribution of $Y^2$ will still tend to a Gaussian in the large-$k$ limit, with different values for $M$ and $V$. With the assumptions in the OP one has $M={\cal O}(k)$ and $V={\cal O}(k)$, so the scaling obtained above still holds: $\mathbb{E}[Y]={\cal O}(k^{1/2})$ and $\text{var}\,[Y]={\cal O}(k^0)$.