Suppose
\begin{align*}
Z_1, \cdots, Z_n \sim \mathcal{N}(0, 1)
\end{align*}
let
\begin{align*}
X = Z_1^2 + \cdots + Z_n^2
\end{align*}
then, X follows a chi-squared distribution \chi_n^2(x) with n degrees of freedom.
\begin{align*}
\chi_n^2(x) &= \frac{1}{2^{\frac{n}{2}} \Gamma(\frac{n}{2})} x^{\frac{n}{2}-1} e^{-\frac{x}{2}} \\
\end{align*}
where x > 0, when x \le 0, \chi_n^2(x) = 0.
Let's use mathematical induction to derive a chi-squared distribution with n dgrees of freedom.
When n=1, X = Z_1^2, z_1 = \sqrt{x}, dz_1 = \frac{1}{2} x^{-\frac{1}{2}} dx, z_1: 0 \rightarrow \infty.
\begin{align*}
1 = \int_0^{\infty} f_X(x) dx &= \int_{-\infty}^{\infty} f_{Z_1}(z_1) d_1z \\
&= 2 \int_{0}^{\infty} f_{Z_1}(z_1) dz_1 \\
&= 2 \int_{0}^{\infty} f_{Z_1}(\sqrt{x}) \frac{1}{2} x^{-\frac{1}{2}} dx \\
&= 2 \int_{0}^{\infty} \frac{1}{\sqrt{2 \pi}} \exp \left(
-\frac{1}{2} (\sqrt{x})^2
\right)\frac{1}{2} x^{-\frac{1}{2}} dx \\
&= \int_{0}^{\infty}
\frac{1}{\sqrt{2 \pi}}
x^{-\frac{1}{2}}
\exp \left(
-\frac{1}{2} x
\right)
dx \\
&= \int_{0}^{\infty}
\frac{1}{2^{\frac{1}{2}}
\Gamma(\frac{1}{2})}
x^{\frac{1}{2}-1}
e^{-\frac{x}{2}}
dx \\
&= \int_{0}^{\infty}
\chi_1^2(x)
dx \\
\end{align*}
therefore
\begin{align*}
f_X(x) = \chi_1^2(x)
\end{align*}
When n=2, let
X = Z_1^2 + Z_2^2 = Y_1 + Y_2
, y_2 = x - y_1
, y_2: 0 \rightarrow x,
, dy_2 = dx
\begin{align*}
1 = \int_0^{\infty} f_X(x) dx &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f_{Z_1 Z_2}(z_1, z_2) dz_1 dz_2 \\
&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f_{Z_1}(z_1) f_{Z_2}(z_2) dz_1 dz_2 \\
&= \int_{0}^{\infty}\int_{0}^{\infty} \chi_1^2(y_1) \chi_1^2(y_2) dy_1 dy_2 \\
&= \int_{0}^{\infty}\int_{0}^{x} \chi_1^2(y_1) \chi_1^2(x - y_1) dy_1 dx \\
&= \int_{0}^{\infty}\int_{0}^{x}
\frac{1}{2^{\frac{1}{2}}
\Gamma(\frac{1}{2})}
y_1^{\frac{1}{2}-1}
e^{-\frac{y_1}{2}}
\frac{1}{2^{\frac{1}{2}}
\Gamma(\frac{1}{2})}
(x - y_1)^{\frac{1}{2}-1}
e^{-\frac{x - y_1}{2}}
dy_1 dx \\
&= \int_{0}^{\infty}
\frac{1}{2 \pi}
e^{-\frac{x}{2}}
\int_{0}^{x}
y_1^{-\frac{1}{2}}
(x - y_1)^{- \frac{1}{2}}
dy_1 dx \\
\end{align*}
Here let y_1 = ux
, u: 0 \rightarrow 1
, dy_1 = x du
\begin{align*}
&= \int_{0}^{\infty}
\frac{1}{2 \pi}
e^{-\frac{x}{2}}
\int_{0}^{1}
(ux)^{-\frac{1}{2}}
(x - ux)^{- \frac{1}{2}}
x du dx \\
&= \int_{0}^{\infty}
\frac{1}{2 \pi}
e^{-\frac{x}{2}}
\int_{0}^{1}
u^{-\frac{1}{2}}
(1 - u)^{- \frac{1}{2}}
du dx \\
&= \int_{0}^{\infty}
\frac{1}{2 \pi}
e^{-\frac{x}{2}}
\Beta(\frac{1}{2}, \frac{1}{2})
dx \\
&= \int_{0}^{\infty}
\frac{1}{2}
e^{-\frac{x}{2}}
dx \\
&= \int_{0}^{\infty}
\chi_2^2(x)
dx \\
\end{align*}
therefore
\begin{align*}
f_X(x) = \chi_2^2(x)
\end{align*}
Suppose a hypothesis,
\begin{align*}
f_X(x) = \chi_k^2(x)
\end{align*}
when n=k.
When n=k+1, let
X = Z_1^2 + \cdots + Z_{k+1}^2 = Y_1 + \cdots + Y_{k+1}
, y = y_1 + \cdots + y_k
, y_{k+1} = x - y
, y_{k+1}: 0 \rightarrow x,
, dy_{k+1} = dx
\begin{align*}
1 = \int_0^{\infty} f_X(x) dx
&= \int_{0}^{\infty}\int_{0}^{x}
\chi_k^2(y) \chi_1^2(x - y) dydx \\
&= \int_{0}^{\infty}\int_{0}^{x}
\frac{1}{2^{\frac{k}{2}} \Gamma(\frac{k}{2})} y^{\frac{k}{2}-1} e^{-\frac{y}{2}}
\frac{1}{2^{\frac{1}{2}} \Gamma(\frac{1}{2})} (x-y)^{-\frac{1}{2}} e^{-\frac{x-y}{2}}
dydx \\
&= \int_{0}^{\infty}\int_{0}^{x}
\frac{1}{2^{\frac{k}{2}} \Gamma(\frac{k}{2})} y^{\frac{k}{2}-1}
\frac{1}{2^{\frac{1}{2}} \Gamma(\frac{1}{2})} (x-y)^{-\frac{1}{2}} e^{-\frac{x}{2}}
dydx \\
&= \int_{0}^{\infty}
\frac{1}{2^{\frac{k+1}{2}} \Gamma(\frac{k}{2})\Gamma(\frac{1}{2})}
e^{-\frac{x}{2}}
\int_{0}^{x}
y^{\frac{k}{2}-1}
(x-y)^{-\frac{1}{2}}
dydx \\
\end{align*}
Here let y = ux
, u: 0 \rightarrow 1
, dy = x du
\begin{align*}
&= \int_{0}^{\infty}
\frac{1}{2^{\frac{k+1}{2}} \Gamma(\frac{k}{2})\Gamma(\frac{1}{2})}
e^{-\frac{x}{2}}
\int_{0}^{x}
(ux)^{\frac{k}{2}-1}
(x-ux)^{-\frac{1}{2}}
x du dx \\
&= \int_{0}^{\infty}
\frac{1}{2^{\frac{k+1}{2}} \Gamma(\frac{k}{2})\Gamma(\frac{1}{2})}
e^{-\frac{x}{2}}
x^{\frac{k+1}{2}}
\int_{0}^{1}
u^{\frac{k}{2}-1}
(1-u)^{\frac{1}{2} - 1}
du dx \\
&= \int_{0}^{\infty}
\frac{1}{2^{\frac{k+1}{2}} \Gamma(\frac{k}{2})\Gamma(\frac{1}{2})}
e^{-\frac{x}{2}}
x^{\frac{k+1}{2}}
\Beta(\frac{k}{2}, \frac{1}{2})
dx \\
&= \int_{0}^{\infty}
\frac{1}{2^{\frac{k+1}{2}} \Gamma(\frac{k}{2})\Gamma(\frac{1}{2})}
e^{-\frac{x}{2}}
x^{\frac{k+1}{2}}
\frac{\Gamma(\frac{k}{2}) \Gamma(\frac{1}{2})}{\Gamma(\frac{k+1}{2})}
dx \\
&= \int_{0}^{\infty}
\frac{1}{2^{\frac{k+1}{2}} \Gamma(\frac{k+1}{2})}
e^{-\frac{x}{2}}
x^{\frac{k+1}{2}}
dx \\
&= \int_{0}^{\infty}
\chi_{k+1}^2(x)
dx \\
\end{align*}
Therefore, the hypothesis also holds when n = k + 1.
Next, let's think about mean and variance of the chi-squared distribution.
The moment generating function of the chi-squared distribution is,
\begin{align*}
M(\theta) &= E[e^{\theta X}] \\
&= \int e^{\theta x} \chi_n^2(x) dx \\
&= \int e^{\theta x} \frac{1}{2^{\frac{n}{2}} \Gamma(\frac{n}{2})} x^{\frac{n}{2}-1} e^{-\frac{x}{2}} dx \\
&= \frac{1}{2^{\frac{n}{2}} \Gamma(\frac{n}{2})} \int x^{\frac{n}{2}-1} e^{- (\frac{1}{2} - \theta) x } dx \\
\end{align*}
let (\frac{1}{2} - \theta)x = t, x = (\frac{1}{2} - \theta)^{-1} t , dx = (\frac{1}{2} - \theta)^{-1} dt, t: 0 \rightarrow \infty.
\begin{align*}
&= \frac{1}{2^{\frac{n}{2}} \Gamma(\frac{n}{2})}
\int_0^\infty
((\frac{1}{2} - \theta)^{-1} t)^{\frac{n}{2}-1} e^{- t } (\frac{1}{2} - \theta)^{-1} dt\\
&= \frac{1}{2^{\frac{n}{2}} \Gamma(\frac{n}{2})}
\frac{1}{(\frac{1}{2} - \theta)^{\frac{n}{2}}}
\int_0^\infty
t^{\frac{n}{2}-1} e^{- t } dt\\
&= \frac{1}{2^{\frac{n}{2}} \Gamma(\frac{n}{2})}
\frac{1}{(\frac{1}{2} - \theta)^{\frac{n}{2}}}
\Gamma(\frac{n}{2})\\
&= \frac{1}{2^{\frac{n}{2}} (\frac{1}{2} - \theta)^{\frac{n}{2}}} \\
&= (1 - 2\theta)^{- \frac{n}{2}} \\
\end{align*}
\begin{align*}
M'(\theta) &= - \frac{n}{2}(1 - 2\theta)^{- \frac{n}{2} - 1} (-2) \\
&= n (1 - 2\theta)^{- \frac{n}{2} - 1} \\
M''(\theta) &= n (- \frac{n}{2} - 1) (1 - 2\theta)^{- \frac{n}{2} - 2} (-2)\\
&= (n^2 + 2n)(1 - 2\theta)^{- \frac{n}{2} - 2}
\end{align*}
The mean and variance are
\begin{align*}
E[X] &= M'(0) = n \\
V[X] &= M''(0) - M'(0)^2 \\
&= (n^2 + 2n) - n^2 \\
&= 2n \\
\end{align*}
Discussion