#### PROOF THAT $$\chi^2_{(1\,\text{df})}$$ MODELS $$X^2$$ WITH $$X\sim N(0,1)$$:

[What follows was first written in a post in SE Mathematics.]

Let’s say that $$X \sim N(0,1)$$ and that $$Y=X^2$$ and find the density of $$Y$$ by using the $$\text{cdf}$$ method:

$$p(Y \leq y) = p(X^2 \leq y)= p(-\sqrt{y} \leq x \leq \sqrt{y})$$. The problem is that we cannot integrate in close form the density of the normal distribution. But we can express it:

$F_X(y) = F_X(\sqrt{y})- F_X(-\sqrt[]{y}).$ Taking the derivative:

$f_X(y)= F_X'(\sqrt{y})\,\frac{1}{2\sqrt{y}}+ F_X'(\sqrt{-y})\,\frac{1}{2\sqrt{y}}.$

Since the values of the normal $$pdf$$ are symmetrical:

$$\large f_X(y)= F_X'(\sqrt{y})\,\frac{1}{\sqrt{y}}$$. Equating this to the $$pdf$$ of the normal (now the $$x$$ in the $$pdf$$ will be $$\sqrt{y}$$ to be plugged into the $$\large e^{-\frac{x^2}{2}}$$ part of the normal $$pdf$$); and remembering to in include $$\large \frac{1}{\sqrt{y}}$$ at the end:

$\large f_X(y)= F_X'(\sqrt[]{y})\,\frac{1}{\sqrt[]{y}}= \frac{1}{\sqrt{2\pi}}\,e^{-\frac{y}{2}}\, \frac{1}{\sqrt[]{y}}=\frac{1}{\sqrt{2\pi}}\,e^{-\frac{y}{2}}\, y^{\frac{1}{2}- 1}$

Compare to the pdf of the chi square:

$\Large f_X(x)= \frac{1}{2^{\nu/2}\Gamma(\frac{\nu}{2})}e^{\frac{-x}{2}}x^{\frac{\nu}{2}-1}$

Since $$\Gamma(1/2)=\sqrt{\pi}$$, for $$1$$ df, we have derived exactly the $$pdf$$ of the chi square.

There is a youtube video dealing with the proof that the sums of the squares of normally distributed $$n$$ random errors, each one distributed as $$\sim \chi^2(1\text{ df})$$ follows a chi square distribution with $$n$$ degrees of freedom:

So if $$X_1, \ldots, X_n$$ are $$\sim \chi^2 \,\small(1\text{ df})$$, then $$X_1 +\cdots+X_n \sim \chi^2\,\small(n\text{ df})$$.

#### PROOF THAT $$X_1, \ldots, X_n$$ ARE $$\sim \chi^2_{(1\,\text{df})}$$, then $$X_1 +\cdots+X_n \sim \chi^2_{(n\,\text{df})}$$:

In probability theory and statistics, the chi-squared distribution (also chi-square or $$\chi^2$$-distribution) with $$k$$ degrees of freedom is the distribution of a sum of the squares of $$k$$ independent standard normal random variables.

Proof through the $$cdf$$:

$$p\,(X_1 +\cdots+X_n \leq x)= p\,(Y_1^2 +\cdots+Y_n^2 \leq x)$$, where $$X_i =Y _i^2$$, and $$Y \sim N(0,1)$$.

$$Y_1^2 +\cdots+Y_n^2 \leq x$$ is an $$n$$-dimensional sphere with radius $$\,\sqrt{x}$$: $$Y_1^2 +\cdots+Y_n^2 \leq \sqrt[]{x^2}$$.

So the probability is:

$$p\,(X_1 +\cdots+X_n \leq x) = \displaystyle \int\cdots\int_{(Y_1^2 +\cdots+Y_n^2 \leq x)}\large f_{\small(Y_1,\ldots,Y_n)}\large(y_1,\ldots,y_n)\, dy_1,\ldots,dy_n$$ with $$f$$ being the joint density. Remembering that they are iid rv’s and that $$Y$$ is normal:

$p(X_1 +\cdots+X_n \leq x) = \displaystyle \int\cdots\int_{(Y_1^2 +\cdots+Y_n^2 \leq x)} \bigg(\frac{1}{2\pi}\bigg)^n e^{-\bigg(\frac{y_1^2}{2}+\cdots + \frac{y_n^2}{2}\bigg) } dy_1\cdots dy_n$

Changing to spherical coordinates it will be something like:

$$dy_1\cdots dy_n=dr\,r\,d\theta_1\,\ldots\,r\,d\theta_n \cdot g(\theta_1\,\ldots\,\theta_n)$$ with $$g$$ being some function of the angles.

And, $\small p(X_1 +\cdots+X_n \leq x) = \displaystyle \int_{r=0}^{\sqrt{x}} \int_{\theta_1=\alpha_1}^{\beta_1} \ldots\int_{\theta_{n-1}=\alpha_{n-1}}^{\beta_{n-1}} \bigg(\frac{1}{2\pi}\bigg)^{n/2}\, e^{-(\frac{r^2}{2}) } \,r^{n-1} \,g(\theta_1\,\ldots\,\theta_{n-1})\, dr\,d\theta_1\cdots \,d\theta_n.$

There is no $$r$$ dependency in the $$\theta$$ integration. The integration of the $$\theta$$’s gets absorbed into a constant that depends on $$n$$, $$C_n$$:

$\Large p(X_1 +\cdots+X_n \leq x) = C_n \cdot \displaystyle \int_{r=0}^{\sqrt{x}} e^{-\bigg(\frac{r^2}{2}\bigg) } r^{n-1} \, dr.$

This is the distribution ($$cdf$$), and to get the density ($$pdf$$) we have to take derivatives:

Using Leibniz rule:

Given $$F(\alpha)=\int_{a(\alpha)}^{b(\alpha)}\,f(x)\,dx$$

$$\frac{d}{d\alpha}\,F(\alpha)=f(b(\alpha))\frac{d}{d\alpha}b(\alpha)\,-\,f(a(\alpha))\frac{d}{d\alpha}b(\alpha)$$

So,

$$\frac{d}{dx}\left[C_n \displaystyle \int_{r=0}^{\sqrt{x}} e^{-\bigg(\frac{r^2}{2}\bigg) } r^{n-1} dr\right] =C_n \, e^{-(\frac{(\sqrt{x})^2}{2})}\,(\sqrt{x})^{n-1}\,1/2(x)^{-1/2}\,-\,C_n\,e^{-0^2/2}\,0^{n-1}\,0$$

$$\large =C_n \frac{1}{2\sqrt{x}}e^{-x/2}x^{(n-1)/2} =\large C_n \frac{1}{2}\frac{1}{\sqrt{x}}\,e^{-x/2}\,x^{1/2(n-1)} =\large C_n e^{-x/2}\,x^{\frac{n-1}{2}}\,\frac{1}{2}\,x^{-\frac{1}{2}}= \large \frac{C_n}{2}\cdot e^{-x/2}\,x^{\frac{n}{2}-1}$$

which is a chi square with n degrees of freedom if we equate $$\frac{C_n}{2}=\frac{1}{2^{n/2}\Gamma(\frac{n}{2})}$$ given that the pdf of chi-square is:

$\displaystyle f_X(x)= \frac{1}{2^{\nu/2}\,\Gamma(\frac{\nu}{2})}\,e^{-x/2}\,x^{\frac{\nu}{2}-1}$ for $$x\leq 0$$. Otherwise, $$0$$.

Geometrically, this is equivalent to the surface area of the $$n$$-dimensional sphere.