# Central Limit Theorem for square roots of sums of i.i.d. random variables

Intrigued by a question at math.stackexchange, and investigating it empirically, I am wondering about the following statement on the square-root of sums of i.i.d. random variables.

Suppose $X_1, X_2, \ldots, X_n$ are i.i.d. random variables with finite non-zero mean $\mu$ and variance $\sigma^2$, and $\displaystyle Y=\sum_{i=1}^n X_i$. The central limit theorem says $\displaystyle \dfrac{Y - n\mu}{\sqrt{n\sigma^2}} \ \xrightarrow{d}\ N(0,1)$ as $n$ increases.

If $Z=\sqrt{|Y|}$, can I also say something like $\displaystyle \dfrac{Z - \sqrt{n |\mu|-\tfrac{\sigma^2}{4|\mu|}}}{\sqrt{\tfrac{\sigma^2}{4|\mu|}}}\ \xrightarrow{d}\ N(0,1)$ as $n$ increases?

For example, suppose the $X_i$ are Bernoulli with mean $p$ and variance $p(1-p)$, then $Y$ is binomial and I can simulate this in R, say with $p=\frac13$:

set.seed(1)
cases <- 100000
n <- 1000
p <- 1/3
Y <- rbinom(cases, size=n, prob=p)
Z <- sqrt(abs(Y))


which gives approximately the hoped-for mean and variance for $Z$

> c(mean(Z), sqrt(n*p - (1-p)/4))
[1] 18.25229 18.25285
> c(var(Z), (1-p)/4)
[1] 0.1680012 0.1666667


and a Q-Q plot which looks close to Gaussian

qqnorm(Z)


• @MichaelM: Thanks for those comments. I had started with the $X_i$ non-negative, but I thought the intuitive asymptotic behaviour you describe allowed a generalisation to more distributions. My surprises were (a) the variance of the square root of the sum apparently tending to a constant not depending on $n$ and (b) the appearance of a distribution which looks very close to Gaussian. A counter-example would be welcome, but when I tried other cases which initially seemed non-Gaussian, increasing $n$ further seemed to bring the distribution back to a CLT-type result. – Henry Oct 21 '16 at 12:00
• A corollary of this is the root-mean-square (or quadratic mean) of i.i.d. random variables suitably scaled (multiply by $\sqrt{n}$ as with an arithmetic mean) also converges to a Gaussian distribution provided that the $4$th moment of the underlying distribution is finite. – Henry Oct 26 '16 at 20:13
• Just a short comment: the claim is a special case of the Delta method, see Theorem 5.5.24 in the book "Statistical inference" by Casella & Berger. – Michael M Oct 30 '16 at 10:34
• @Michael: Perhaps you see something that I am not at the moment, but I do not think this particular problem fits within the assumptions of the classical Delta method (e.g., as stated in the theorem you reference). Note that $Y$ does not converge in distribution (nontrivially on $\mathbb R$) and so "applying the Delta method with $g(y) = \sqrt{|y|}$" does not satisfy the requisite requirements. However, as S. Catterall's answer demonstrates, it provides a useful heuristic which leads to the correct answer. – cardinal Jul 22 '17 at 21:13
• (I believe you could adapt the proof of the Delta method to cases similar to the above in order to make fully rigorous the aforementioned heuristic.) – cardinal Jul 22 '17 at 21:15

Suppose that $X_1,X_2,X_3,...$ are IID random variables with mean $\mu\gt 0$ and variance $\sigma^2$, and define the sums $Y_n=\sum_{i=1}^n X_i$. Fix a number $\alpha$. The usual Central Limit Theorem tells us that $P(\frac{Y_n-n\mu}{\sigma\sqrt n}\leq \alpha)\to\Phi(\alpha)$ as $n\to\infty$, where $\Phi$ is the standard normal cdf. However, the continuity of the limiting cdf implies that we also have $$P\Big(\frac{Y_n-n\mu}{\sigma\sqrt n}\leq \alpha+\frac{\alpha^2 \sigma^2}{4\mu\sigma\sqrt n}\Big)\to\Phi(\alpha)$$ because the additional term on the right hand side of the inequality tends to zero. Rearranging this expression leads to $$P\Big(Y_n\leq (\frac{\alpha\sigma}{2\sqrt \mu}+\sqrt{n\mu})^2\Big)\to\Phi(\alpha)$$
Taking square roots, and noting that $\mu\gt 0$ implies that $P(Y_n\lt 0)\to 0$, we obtain $$P\Big(\sqrt{|Y_n|}\leq \frac{\alpha\sigma}{2\sqrt \mu}+\sqrt{n\mu}\Big)\to\Phi(\alpha)$$ In other words, $\frac{\sqrt{|Y_n|}-\sqrt{n\mu}}{\sigma/{2\sqrt\mu}}\xrightarrow{d}N(0,1)$. This result demonstrates convergence to a Gaussian in the limit as $n\to\infty$.
Does this mean that $\sqrt{n\mu}$ is a good approximation to $E[\sqrt{|Y_n|}]$ for large $n$? Well, we can do better than this. As @Henry notes, assuming everything is positive, we can use $E[\sqrt{Y_n}]=\sqrt{E[Y_n]-\text{Var}(\sqrt{Y_n})}$, together with $E[Y_n]=n\mu$ and the approximation $\text{Var}(\sqrt{Y_n})\approx \frac{\sigma^2}{4\mu}$, to obtain the improved approximation $E[\sqrt{|Y_n|}]\approx\sqrt{n\mu- \dfrac{\sigma^2}{4\mu}}$ as stated in the question above. Note also that we still have $$\frac{\sqrt{|Y_n|}-\sqrt{n\mu-\frac{\sigma^2}{4\mu}}}{\sigma/{2\sqrt\mu}}\xrightarrow{d}N(0,1)$$ because $\sqrt{n\mu-\frac{\sigma^2}{4\mu}}-\sqrt{n\mu}\to 0$ as $n\to\infty$.
• You may need to add $\sqrt{n \mu}-\sqrt{n \mu-\tfrac{\sigma^2}{4\mu}} \to 0$ as ${n \to \infty}$ to get my result – Henry Oct 24 '16 at 21:44
• @Henry You can replace $\sqrt{n\mu}$ with $\sqrt{n\mu-k}$ for any constant $k$ and this won't change the limiting distribution, but it may change the degree to which $\frac{\sqrt{|Y_n|}-\sqrt{n\mu-k}}{\sigma/{2\sqrt\mu}}$ is a good approximation to $N(0,1)$ for specific large $n$ . How did you come up with $\sqrt{n \mu-\tfrac{\sigma^2}{4\mu}}$? – S. Catterall Reinstate Monica Oct 25 '16 at 9:44
• We have $\text{Var}(Z)=E[Z^2]-(E[Z])^2$ so $E[Z]=\sqrt{E[Z^2]-\text{Var}(Z)}$. Assuming everything is positive, $E[Z^2]=E[Y]=n\mu$ while the denominator of $\frac{\sqrt{|Y_n|}-\sqrt{n\mu}}{\sigma/{2\sqrt\mu}}$ suggests $\text{Var}(Z) \approx \dfrac{\sigma^2}{4\mu}$, and combining these leads to $E[Z] \approx \sqrt{n\mu- \dfrac{\sigma^2}{4\mu}}$. – Henry Oct 25 '16 at 14:51