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Such a result can be obtained by the so-called delta method, which yields, in particular, the following: if $X_m\sim\chi^2_m$, then the distribution of $\sqrt X_m$ is approximately $N(\sqrt m,1/\sqrt2)$ (for large $m$).

Details: By the central limit theorem, $\bar X_m:=X_m/m$ is approximately normal with mean $\mu:=1$ and standard deviation $\sqrt{2/m}$. Hence, by the delta method, for $g(x)\equiv\sqrt x$, $\sqrt{\bar X_m}$ is approximately normal with mean $g(\mu)=\sqrt1=1$ and standard deviation $g'(\mu)\sqrt{2/m}=1/\sqrt{2m}$. Thus, $\sqrt X_m=\sqrt m\,\sqrt{\bar X_m}$ is approximately normal with mean $\sqrt m$ and standard deviation $1/\sqrt2$, as claimed.

On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly) multivariate delta method, see this and references there.

The convergence of the distribution of $\sqrt X_m$ to normality is in a sense monotonic; see formula (2.6).

Such a result can be obtained by the so-called delta method, which yields, in particular, the following: if $X_m\sim\chi^2_m$, then the distribution of $\sqrt X_m$ is approximately $N(\sqrt m,1/\sqrt2)$ (for large $m$).

Details: By the central limit theorem, $\overline X_m:=X_m/m$ is approximately normal with mean $\mu:=1$ and standard deviation $\sqrt{2/m}$. Hence, by the delta method, for $g(x)\equiv\sqrt x$, $\sqrt{\overline X_m}$ is approximately normal with mean $g(\mu)=\sqrt1=1$ and standard deviation $|g'(\mu)|\sqrt{2/m}=1/\sqrt{2m}$. Thus, $\sqrt X_m=\sqrt m\,\sqrt{\overline X_m}$ is approximately normal with mean $\sqrt m$ and standard deviation $1/\sqrt2$, as claimed.

On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly multivariate) delta method, see this and references there.

The convergence of the distribution of $\sqrt X_m$ to normality is in a sense monotonic; cf. formula (2.6).

Such a result can be obtained by the so-called delta method, which yields, in particular, the following: if $X_m\sim\chi^2_m$, then the distribution of $\sqrt X_m$ is approximately $N(\sqrt m,1/\sqrt2)$ (for large $m$).

Details: By the central limit theorem, $\bar X_m:=X_m/m$ is approximately normal with mean $\mu:=1$ and standard deviation $\sqrt{2/m}$. Hence, by the delta method, for $g(x)\equiv\sqrt x$, $\sqrt{\bar X_m}$ is approximately normal with mean $g(\mu)=\sqrt1=1$ and standard deviation $g'(\mu)\sqrt{2/m}=1/\sqrt{2m}$. Thus, $\sqrt X_m=\sqrt m\,\sqrt{\bar X_m}$ is approximately normal with mean $\sqrt m$ and standard deviation $1/\sqrt2$, as claimed.

On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly) multivariate delta method, see this and references there.

Such a result can be obtained by the so-called delta method, which yields, in particular, the following: if $X_m\sim\chi^2_m$, then the distribution of $\sqrt X_m$ is approximately $N(\sqrt m,1/\sqrt2)$ (for large $m$).

Details: By the central limit theorem, $\bar X_m:=X_m/m$ is approximately normal with mean $\mu:=1$ and standard deviation $\sqrt{2/m}$. Hence, by the delta method, for $g(x)\equiv\sqrt x$, $\sqrt{\bar X_m}$ is approximately normal with mean $g(\mu)=\sqrt1=1$ and standard deviation $g'(\mu)\sqrt{2/m}=1/\sqrt{2m}$. Thus, $\sqrt X_m=\sqrt m\,\sqrt{\bar X_m}$ is approximately normal with mean $\sqrt m$ and standard deviation $1/\sqrt2$, as claimed.

On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly) multivariate delta method, see this and references there.

The convergence of the distribution of $\sqrt X_m$ to normality is in a sense monotonic; see formula (2.6).

Such a result can be obtained by the so-called delta method, which yields, in particular, the following: if $X_m\sim\chi^2_m$, then the distribution of $\sqrt X_m$ is approximately $N(\sqrt m,1/\sqrt2)$ for large $m$.

Such a result can be obtained by the so-called delta method, which yields, in particular, the following: if $X_m\sim\chi^2_m$, then the distribution of $\sqrt X_m$ is approximately $N(\sqrt m,1/\sqrt2)$ (for large $m$).

Details: By the central limit theorem, $\bar X_m:=X_m/m$ is approximately normal with mean $\mu:=1$ and standard deviation $\sqrt{2/m}$. Hence, by the delta method, for $g(x)\equiv\sqrt x$, $\sqrt{\bar X_m}$ is approximately normal with mean $g(\mu)=\sqrt1=1$ and standard deviation $g'(\mu)\sqrt{2/m}=1/\sqrt{2m}$. Thus, $\sqrt X_m=\sqrt m\,\sqrt{\bar X_m}$ is approximately normal with mean $\sqrt m$ and standard deviation $1/\sqrt2$, as claimed.

On the uniform and nonuniform bounds on the rate of convergence to normality in the (possibly) multivariate delta method, see this and references there.