Proofs related to chi-squared distribution for k degrees of freedom

I was reading a proof related to chi-squared distribution for k degrees of freedom from wiki.

https://en.wikipedia.org/wiki/Proofs_related_to_chi-squared_distribution

I think I might understand the general idea behind the proof. But there are some subtle details which I am confused about.

1) What is the meaning of the notation $P(Q)dQ$? Shouldn't it just be $P(Q)$?

2)The integral $\int_vdx_1dx_2...dx_k$ is equal to the surface area of the (k − 1)-sphere times the infinitesimal thickness of the sphere which is $dR = dQ/2Q^{1/2}$. Why we need to times $dR$?

Could someone please help me with the questions?

Solution 1:

The proof indeed involves some knowledge of integration.

We want to show that if $X_1,\ldots, X_n$ are independent with law $\mathcal N(0,1)$, then $$U := U_n := X_1^2+\ldots +X_n^2 \sim \chi_n^2,$$ i.e. $U$ admits the density \begin{align*} f_{\chi_n^2}(t) = \frac{t^{n/2-1}}{2^{n/2}\Gamma(n/2)}\, e^{-t/2}\, \mathbb 1_{\mathbb R_+}(t), \quad t\in\mathbb R. \end{align*}

By definition, the density of the joint probability of $(X_1,\ldots, X_n)$ is \begin{align*} \rho(x_1, ..., x_n) = \prod_{i=1}^n f_{X_i}(x_i) = \frac{1}{ (2\pi)^{n/2} \sigma_1 ... \sigma_n } \exp\left( - \sum_{i=1}^n \frac{(x_i-\mu_i)^2}{2 \sigma_i^2} \right), \end{align*} where $\mu_1 = ... = \mu_n = 0$ et $\sigma_1 = ... = \sigma_n = 1$. For all integrable functions $f(x_1, ..., x_n)$ of $n$ variables $x_1, ..., x_n$, we then have \begin{align*} \int_{|x|^2 < r} f(x_1, ..., x_n) \mathrm d x = \int_0^{\sqrt r} \mathrm d s \int_{|x|= s} f(x) \mathrm d S(x). \end{align*} If $f$ is a radial symmetric function, i.e. $f(x) = f(|x|)$, then \begin{align*} \int_{|x| = r} f(x) \mathrm d S(x) = \sigma_{n-1} f(r) r^{n-1}, \end{align*} with $\mathrm d S(x)$ denoting the surface element of the sphere $\{|x| = r\}$ and $\sigma_{n-1} = \frac{2\pi^{n/2}}{\Gamma(n/2)}$ is the area of the $(n-1)$-sphere. Since the density of $U$ is the derivative of the probability law, \begin{align*} f_U(t) = \frac{1}{2\sqrt t} \int_{|x| = \sqrt t} \rho(x) \mathrm d S(x) \end{align*} and we get \begin{align*} \frac{1}{2\sqrt t} \int_{|x| = \sqrt t} \rho(x) \mathrm d S(x) &= \frac{1}{2 \sqrt t} \ \frac{2 \pi^{n/2}}{\Gamma(n/2)} \frac{e^{-t/2}}{(2\pi)^{n/2}} t^{(n-1)/2} \\ &= t^{-1/2} \frac{t^{(n-1)/2}}{2^{n/2}\Gamma(n/2)} e^{-t/2} = \frac{t^{n/2-1}}{2^{n/2}\Gamma(n/2)} e^{-t/2}. \end{align*}