$$\begin{align}
E\left\lbrace e^2(n) \right\rbrace&=E\left\lbrace\left( f(n)-\sum^m_{i=1}{\alpha}_if(n-i) \right)^2\right\rbrace \\
&=E\left\lbrace f(n)^2-2f(n){\sum^m_i{\alpha}_if(n-i)}+\left(\sum^m_i{\alpha}_if(n-i)\right)^2\right\rbrace \\
&\triangleq E\left\lbrace f(n)^2-A+B \right\rbrace,
\end{align}$$ 要使它最小，我们将它分别对 $\alpha_k, k\in\lbrace 1,2,…,m\rbrace$ 求偏导，求当偏导为 $0$ 时 $\alpha_k$ 的值．
$A$ 在展开后与 $\alpha_k$ 有关的只有 $2\alpha_kf(n)f(n-k)$ 一项．因此 $$\begin{align}
\frac{\partial A}{\partial \alpha_k}&=\frac{\partial}{\partial \alpha_k}2\alpha_kf(n)f(n-k) \\
&=2f(n)f(n-k).
\end{align}$$
而对于 $B$，我们有 $$\begin{align}
B&=\left(\sum^m_i{\alpha}_if(n-i)\right)^2 \\
&\triangleq \left(C+{\alpha}_kf(n-k)\right)^2 \\
&=C^2+2C{\alpha}_kf(n-k)+\left({\alpha}_kf(n-k)\right)^2,
\end{align}$$ 其中 $C$ 不包含与 $\alpha_k$ 有关的项．于是 $$\begin{align}
\frac{\partial B}{\partial \alpha_k}&=2Cf(n-k)+2{\alpha}_k\left(f(n-k)\right)^2 \\
&=2f(n-k)\left(C+\alpha_kf(n-k)\right) \\
&=2f(n-k)\sum^m_i\alpha_if(n-i).
\end{align}$$
故 $$\frac{\partial E\left\lbrace e^2(n) \right\rbrace}{\partial \alpha_k}=E\left\lbrace -2f(n)f(n-k)+2f(n-k)\sum^m_i\alpha_if(n-i) \right\rbrace=0,$$ $$ E\left\lbrace f(n)f(n-k)\right\rbrace =E\left\lbrace f(n-k)\sum^m_i\alpha_if(n-i)\right\rbrace,$$ 对 $k\in\lbrace 1,2,…,m\rbrace$，我们有 $$\begin{align}
E\left\lbrace f(n)f(n-1)\right\rbrace &=E\left\lbrace f(n-1)\sum^m_i\alpha_if(n-i)\right\rbrace, \\
E\left\lbrace f(n)f(n-2)\right\rbrace &=E\left\lbrace f(n-2)\sum^m_i\alpha_if(n-i)\right\rbrace, \\
&\vdots \\
E\left\lbrace f(n)f(n-m)\right\rbrace &=E\left\lbrace f(n-m)\sum^m_i\alpha_if(n-i)\right\rbrace, \\
\end{align}$$ 写成矩阵的形式，就是 $\boldsymbol{r}=\boldsymbol{R}\boldsymbol{\alpha}$，其中
$$\begin{align}
\boldsymbol{r}&=\left(\begin{matrix}
E\left\lbrace f(n)f(n-1)\right\rbrace \\
E\left\lbrace f(n)f(n-2)\right\rbrace \\
\vdots \\
E\left\lbrace f(n)f(n-m)\right\rbrace
\end{matrix}\right),\\
\boldsymbol{R}&=\left(\begin{matrix}
E\left\lbrace f(n-1)f(n-1)\right\rbrace & E\left\lbrace f(n-1)f(n-2)\right\rbrace & \cdots & E\left\lbrace f(n-1)f(n-m)\right\rbrace \\
E\left\lbrace f(n-2)f(n-1)\right\rbrace & E\left\lbrace f(n-2)f(n-2)\right\rbrace & \cdots & E\left\lbrace f(n-2)f(n-m)\right\rbrace \\
\vdots & \vdots & & \vdots \\
E\left\lbrace f(n-m)f(n-1)\right\rbrace & E\left\lbrace f(n-m)f(n-2)\right\rbrace & \cdots & E\left\lbrace f(n-m)f(n-m)\right\rbrace \\
\end{matrix}\right),\\
\boldsymbol{\alpha}&=\left(\begin{matrix}
\alpha_1 \\
\alpha_2 \\
\vdots \\
\alpha_m
\end{matrix}\right).
\end{align}$$ 于是 $$\boldsymbol{\alpha}=\boldsymbol{R}^{-1}\boldsymbol{r}.$$ $\boldsymbol{\alpha}$ 可以按此式求解．