Chi-squared convergence to the Gaussian distribution

Question

I'm taking a Statistics course this semester and the professor mentioned that the chi-squared distribution $\chi_n^2$ satisfies that $$\sqrt{2\chi_n^2}-\sqrt{2n-1}$$

converges to a Gaussian distribution. I have been looking for a proof of this result, but haven't found one.

We have defined the chi-squared distribution via its density function:

$$ f(x)=\left\{\begin{array}{rl} 0 & \text{if }x\leq 0 \\ \frac{x^{n/2-1}e^{x/2}}{2^{n/2}\Gamma(n/2)} & \text{if }x>0 \end{array} \right. $$

and also shown that, if n is a positive integer, it follows the same distribution as the sum of $n$ independent squared gaussian variables. I am familiar with the concept of generating functions and the central limit theorem.

Is there a reference for a proof of this result? Preferrably, one that uses the definitions and results stated above. If not, could I be provided with one?

Answer 1

As the comment suggests, the central limit theorem renders this fairly simple to prove, but you will need some background in asymptotic theory for this approach.

Note that since a chi-squared random variable with $n$ degrees of freedom is a sum of $n$ squared independent standard normal random variables, your problem is to show

$$Y_n:=\sqrt{2\chi_n^2}-\sqrt{2n-1}=\sqrt{n} \left( g\left(\frac{1}{n}\sum_{i=1}^n Z_i^2\right)-g\left(1-{1\over 2n}\right)\right) \xrightarrow{d} N(\mu,\sigma^2)$$

for suitable $\mu,\sigma^2$, where $\xrightarrow{d}$ denotes convergence in distribution, $Z_i\sim N(0,1)$ are independent, and $g(x):=\sqrt {2x}$. Writing the problem this way allows us to appeal to some convergence theorems.

First, note that by CLT and Slutsky's theorem,

$$ \sqrt{n} \left(\left(\frac{1}{n}\sum_{i=1}^n Z_i^2\right)-\left(1-{1\over 2n}\right)\right)= \underbrace{\sqrt{n} \left(\left(\frac{1}{n}\sum_{i=1}^n Z_i^2\right)-1\right)}_{\xrightarrow{d} N(0,2)}+\underbrace{{1 \over 2\sqrt n}}_{\xrightarrow{d} 0} \xrightarrow{d}N(0,2).$$

Next, the delta method tells us

$$Y_n \xrightarrow{d} g'(1)N(0,2)=\frac{1}{\sqrt 2}N(0,2)=N(0,1),$$

and we are done.

Answer 2

Starting with the random variable $X= \chi_n^2$ whose known probability density function is $dP= f_X(x)dx \sim x^{(\frac {n}{2}-1)} e^{-x}$ as described in your question, pass to the new random variable $Y =\sqrt{2 X }$ whose pdf can be deduced from the formal requirement $dP= f_Y(y) dy= f_X(x) dx$. After you make the substitution $x= y^2/2$ to change variables, you should get (modulo a harmless constant factor) something like $f_Y(y)dy= e^{-y^2/4} y^{n-1} dy$. At this stage you can learn a lot by plotting the graph of $f_Y(y) dy$ for various values of $n$.

Digging deeper. You can make use of a standard technique in asymptotic approximations called the Laplace/Watson Method to get insight into the behavior of the density near its peak. Write the density as $f_Y(y) dy=e^{ h(y)} dy $ where $h(y)=-y^2/4 + (n-1)\ln (y)$. Note that $h(y)$ has only one positive critical point, located at $y=c =\sqrt{2n-1}$ and looks like a quadratic in the vicinity of that peak location. (In fact $h(y-c)= -w^2$ for some smooth function $w(y)$ that is well-behaved near $y=c$, a special case of the Morse Lemma.) Thus the shifted random variable $Z=Y-c$ has the pdf $ f_Z(z)dz= e^{- h(z) } dz= e^{- h(y-c)} dy = e^{- w^2} \frac{ dy}{dw} dw$ which strongly resembles a Gaussian since $\frac{dy}{dw}$ is approximately constant in a region near the critical point at $w=0$. You can use this idea to estimate the expectation of any interval that is reasonably close to the central hump, and get estimates for how it depends on $n$.

This quick sketch points you toward the main ideas and might help you see intuitively what's happening as $n$ increases.