How can one proof the following:

Let SnS_n be χ2n\chi_n^2-distributed. Then √Sn−√n\sqrt{S_n}-\sqrt{n} is asymptotically N(0,12)N(0,\frac{1}{2}) distributed, i.e. √Sn−√n\sqrt{S_n}-\sqrt{n} ~. N(0,12)N(0,\frac{1}{2}).

I wanted to use the follwing result: If √n(Tn−n)\sqrt{n}(T_n-n) ~. N(0,σ2)N(0, \sigma^2), then for a differentiable function hh with h(μ)≠0h(\mu) \neq 0 we get

√n(h(Tn)−h(μ))\sqrt{n}(h(T_n)-h(\mu)) ~. N(0,σ2h′(μ)2),N(0,\sigma^2 h'(\mu)^2),

but I don’t know how to start. Can someone give me hint or help me?

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would it help that SnS_n ~Gamma(n/2,1/2) Gamma(n/2,1/2)? That can be proved using the mgf of the gamma distribution after having found that the square of a standard normal is gamma.

– Max Freiburghaus

2 days ago

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1 Answer

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We have fSn(x)=xn/2−1e−x/22n/2Γ(n/2),x>0.f_{S_n}(x) = \frac{x^{n/2-1} e^{-x/2}}{2^{n/2} \Gamma(n/2)}, \quad x > 0. Thus, Y=√Sn−√nY = \sqrt{S_n} – \sqrt{n} has density fY(y)=fSn((y+√n)2)2(y+√n)=(y+√n)n−1e−(y+√n)2/22n/2−1Γ(n/2),y>−√n.f_Y(y) = f_{S_n}((y+\sqrt{n})^2) 2(y+\sqrt{n}) = \frac{(y+\sqrt{n})^{n-1} e^{-(y+\sqrt{n})^2/2}}{2^{n/2-1} \Gamma(n/2)}, \quad y > -\sqrt{n}. Using Stirling’s approximation for Γ(n/2)∼√π√n−1/2(n/2−1e)n/2−1+O(1/n)\Gamma(n/2) \sim \sqrt{\pi} \sqrt{n – 1/2} \left(\frac{n/2-1}{e}\right)^{n/2-1} + O(1/n) and taking the logarithm, we get for n>2n > 2 log(√πfY(y))≈n−12log(y+√n)2n−2−y22−y√n−1,\log (\sqrt{\pi} f_Y(y)) \approx \frac{n-1}{2} \log \frac{(y + \sqrt{n})^2}{n-2} – \frac{y^2}{2} – y \sqrt{n} -1, the limit of which as n→∞n \to \infty is −y2-y^2, by a straightforward calculation (e.g. L’Hopital’s rule). Therefore, as n→∞n \to \infty we have fY(y)→e−y2/√π,f_Y(y) \to e^{-y^2}/\sqrt{\pi}, which is normal with mean 00 and variance 1/21/2.