Using CLT, Slutsky's theorem and delta method

Question

Let $Y_n$ be a sequence of random variables with $\chi^2_n$ distribution. Using Slutsky' theorem or delta method prove that $$\sqrt{2Y_n}-\sqrt{2n-1}\stackrel{D}\to N(0,1)$$ In the first place I proved from CLT that $\frac{Y_n-n}{\sqrt{2n}}\stackrel{D}\to N(0,1)$, but I have no idea how to prove this upper convergence using Slutsky's theorem or delta method. My attempt was to write it down as $\frac{2Y_n-(2n-1)}{\sqrt{2Y_n}+\sqrt{2n-1}}$, but I can't see a point here.

Answer 1

From the CLT, $$\sqrt{n}\left(\frac{Y_n}{n} - 1\right) \overset{d}{\to} N(0,2).$$

With $g(u) = \sqrt{u}$, the delta method implies $$\sqrt{n}\left(g\left(\frac{Y_n}{n}\right) - g(1)\right) \overset{d}{\to} N(0, 2(g'(1))^2).$$

Simplifying yields $$\sqrt{2Y_n} - \sqrt{2n} \overset{d}{\to} N(0, 1).$$ Finally apply Slutsky's theorem with the fact that $\sqrt{2n}-\sqrt{2n-1} = \frac{1}{\sqrt{2n}+\sqrt{2n-1}} \to 0$.

Answer 2

As the statement is about distributions, we assume that $Y_{n}=\sum_{k=1}^{n}Z_{k}^{2}$ for some iid $N(0,1)$ variates defined on some fixed probability space.

$$\frac{2Y_{n}-(2n-1)}{\sqrt{2Y_{n}}+\sqrt{2n-1}}=\frac{2Y_{n}-2n}{\sqrt{2n-1}}\cdot\frac{\sqrt{2n-1}}{\sqrt{2Y_{n}}+\sqrt{2n-1}}$$

Now, note that $\displaystyle\cdot\frac{2Y_{n}-2n}{\sqrt{2n-1}}\xrightarrow{D}2\cdot N(0,1)$ by applying Slutsky's Theorem (by noting that $\frac{\sqrt{2n-1}}{\sqrt{2n}}\to 1$)

Now $$\frac{\sqrt{2n-1}}{\sqrt{2Y_{n}}+\sqrt{2n-1}}=\frac{1}{\sqrt{\frac{2Y_{n}}{2n-1}}+1}$$

Now By the Law of Large Numbers, you have that $$\frac{Y_{n}}{n}=\frac{\sum_{k=1}^{n}Z_{k}^{2}}{n}\xrightarrow{a.s.}1$$

Thus, you have $$\frac{\sqrt{2n-1}}{\sqrt{2Y_{n}}+\sqrt{2n-1}}=\frac{1}{\sqrt{\frac{2Y_{n}}{2n-1}}+1}\xrightarrow{a.s.}\frac{1}{2}$$.

Thus, by Slutsky's Theorem again, i.e. if $X_{n}\xrightarrow{d}X$, $Y_{n}\xrightarrow{P}c$ then $X_{n}Y_{n}\xrightarrow{d}c\cdot X$

you have \begin{align}\sqrt{2Y_n}-\sqrt{2n-1}=&\frac{2Y_{n}-2n}{\sqrt{2n-1}}\cdot\frac{\sqrt{2n-1}}{\sqrt{2Y_{n}}+\sqrt{2n-1}}\to 2\cdot N(0,1)\cdot\frac{1}{2}\\\\&=N(0,1)\end{align}