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?
 A: 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*}
A: I think the $dQ$ simply signifies that it is the probability density at a given point on $Q$ so you would need to integrate across some range of Q to assess the size of some probability.
The height of a probability distribution function represents the probability at any given infinitesimal point so to sum the total probability it is necessary to multiply the area by the height, hence times $dR$.  If we use the discrete analogue of a dice, the height of the function at each point is $1/6$ so to sum the probability across any area we multiply the number of points by the height at each point which is 1/6.
$dR$ is like the height of $1/6$ and $6$ is like the total surface area (in this case 1-dimensional length).
A: 1-) The meaning of ∫P(Q)*dQ is the probability of finding the value of Q, according to the limits of the integration.
2-) The use of dR in the proof is fundamental to simplify the description of the shield of the sphere. Since we are talking about a sphere, nothing more appropriate than use spherical coordinates. In cartesian coordinates, it would be very difficult. Then, we can replace the original variables xi's for R.
As I understood this proof. For a given value of xi's, such that Q=x1^2+x2^2+..+xN^2, the probability density is given by P(x1)P(x2)..P(xN), which is, for this case 1/(2*pi)^N*exp(-Q/2). Such probability density will be the same for all points in the shield of the sphere.
In statistics, the change of variables between y and x is always performed ∫P(y)*dy = ∫P(x)*dx with the correpondent limits of x and y.
So, any P(Q)*dQ should correspond to integrate P(xi's)*dxi's along the shield of the sphere.
Thus, ∫P(Q)*dQ = ∫∫..∫P(x1,x2,...,xN)*dx1.dx2...dxN, being the limits of the integral in xi's the correspondent to the shield of the N-sphere. However, it can be simplified, since P(x1,x2,...,xN)=1/(2*pi)^N*exp(-Q/2), and Q is constant and can go out of the integral. Thus:
∫P(Q)*dQ =1/(2*pi)^Nexp(-Q/2)∫∫..∫dx1.dx2...dxN
But ∫∫..∫dx1.dx2...dxN is nothing more than the volume of the shield, which can be admited as A*dR, where A is the area of N-sphere, and R is the radius, R=Q^0.5.
Then
P(Q)*dQ =1/(2*pi)^N*exp(-Q/2)AdR
The deduction follows as the link of Wikipedia.
https://en.wikipedia.org/wiki/Proofs_related_to_chi-squared_distribution
