Overview
These conclusions provide a theoretical basis for the hypothesis testing of parameters and are very important. Refer to "Probability Theory and Mathematical Statistics" to record it.
important theorem
Let \(x_1,\cdots, x_n\) be a sample from a normal population \(N(\mu, \sigma^2)\) , and its sample mean and sample variance are
\begin{align}
\bar{x } &= \frac{1}{n} \sum_{i=1}^n x_i \\
s^2 &= \frac{1}{n-1} \sum_{i=1}^n(x_i - \bar{x})^2
\end{align}
has
- \(\bar{x}\) and \(s^2\) are independent of each other
- \(\bar{x} \sim N(\mu, \dfrac{\sigma^2}{n})\)
- \(\frac{(n-1)s^2}{\sigma^2}\sim \mathcal{X}^2(n-1)\)
Proof: write \(x=(x_1,\cdots,x_n)^T\) , then
\begin{align}
E(X) = \begin{bmatrix} \mu \\ \vdots \\ \mu \end{ bmatrix}, \quad Var(X) = \sigma^2 I
\end{align}
Take a \(n\)- dimensional orthogonal matrix \(A\) , and each element in the first row is \(1 / \sqrt{n}\) such as
\begin{align}
A = \begin{bmatrix} \dfrac{1}{\sqrt{n}} & \dfrac{1}{\sqrt{n}} & \dfrac {1}{\sqrt{n}} & \cdots & \dfrac{1}{\sqrt{n}} \\
\dfrac{1}{\sqrt{2\cdot 1}} & -\dfrac{1} {\sqrt{2\cdot 1}} & 0 & \cdots & 0 \\
\dfrac{1}{\sqrt{3\cdot 2}} & \dfrac{1}{\sqrt{3 \cdot 2}} & -\dfrac{2}{\sqrt{3\cdot 2}} & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\dfrac{1}{\sqrt{n(n-1)}} & \dfrac{1}{\sqrt{n(n-1)}} & \dfrac{1}{\sqrt{n(n-1 )}} & \cdots & -\dfrac{n-1}{\sqrt{n(n-1)}} \\
\end{bmatrix}
\end{align}
Let \(Y=AX\) , then by The properties of the multi-dimensional normal distribution know that \(Y\) still obeys the \(n\)- dimensional normal distribution, and its mean and variance are
\begin{align}
E(Y) &= A \cdot E(X) = \ begin{bmatrix} \sqrt{n} \mu \\ 0 \\ \vdots \\ 0 \end{bmatrix} \\
Var(Y) &= A\cdot Var(X) \cdot A^T = A \cdot \sigma^2 I \cdot A^T = \sigma^2 AA^T = \sigma^2 I
\end{align}
So the components of \(Y=(y_1,\cdots,y_n)^T\) are mutually Independent, and all obey the normal distribution, their variances are all \(\sigma^2\) , and the mean is not exactly the same, the mean of \(y_1\) is \(\sqrt{n}\mu\) , and \(y_2 ,\cdots, y_n\)The mean is \(0\) . Note that \(\bar{x}=\dfrac{1}{\sqrt{n}}y_1\) , which proves conclusion 2.
Since \(\sum_{i=1}^n y_i^2 = Y^TY = X^TA^TAX=\sum_{i=1}^n x_i^2\) , so
\begin{align}
(n- 1)\cdot s^2 &= \sum_{i=1}^n (x_i-\bar{x})^2 = \sum_{i=1}^n x_i^2 - (\sqrt{n}\ bar{x})^2 \\
&=\sum{i=1}^n y_i^2-y_1^2=\sum_{i=2}^n y_i^2
\end{align}
This proves the conclusion 1.
Since \(y_2,\cdots, y_n\) is independent and identically distributed in \(N(0,\sigma^2)\) , then
\begin{align}
\frac{(n-1)s^2}{\sigma ^2} = \sum_{i=2}^n \left(\frac{y_i}{\sigma} \right)^2 \sim \mathcal{X}^2(n-1)
\end{align}
theorem Proof is complete.
important inferences
Corollary 1: Under the notation of the above theorem, we have:
\begin{align} \label{e1}
t = \frac{\sqrt{n}(\bar{x}-\mu)}{s} \sim t( n-1)
\end{align}
proof: From conclusion 2 of the above theorem:
\begin{align}
\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} = N(0 ,1)
\end{align}
and then rewrite the left end of \ref{e1} to
\begin{align}
\frac{\sqrt{n}(\bar{x}-\mu)}{s} = \dfrac{\ dfrac{\bar{x}-\mu}{\sigma / \sqrt{n}}}{\sqrt{\dfrac{(n-1)\cdot s^2 / \sigma^2}{n-1} }}
\end{align}
Since the numerator is a standard normal variable, the root sign of the denominator is a \(t\) variable with degrees of freedom \(n-1\) divided by its degrees of freedom, and the numerator and denominator are mutually Independence, according to the \(t\) distribution definition \(t \sim t(n-1)\) , the proof is completed. Inference 2: Let \(x_1,x_2,\cdots, x_m\) be from
\(N(\mu_1,\sigma_1)\) samples, \(y_1,y_2,\cdots, y_n\) are samples from \(N(\mu_2,\sigma_2)\) , and the two samples are independent of each other , remember
\begin{align}
s_x^2 = \dfrac{1}{m-1}\sum_{i=1}^m(x_i-\bar{x})^2,\quad s_y^2 = \dfrac {1}{m-1}\sum_{i=1}^n(y_i-\bar{y})^2
\end{align}
where
\begin{align}
\bar{x}= \frac{1} {m} \sum_{i=1}^m x_i, \quad \bar{y} = \frac{1}{n}\sum_{i=1}^n y_i
\end{align}
then
\begin{ align}
F = \frac{s_x^2 / \sigma_1^2}{s_y^2 / \sigma_2^2} \sim F(m-1, n-1)
\end{align}
In particular, if \(\ sigma_1^2 = \sigma_2^2\) , then \(F = s_x^2 / s_y^2 \sim F(m-1,n-1)\) .
Prove: From the two samples independently, \(s_x^ 2\)Independent of \(s_y^2\) , and
\begin{align}
\dfrac{(m-1)s_x^2}{\sigma_1^2} \sim \mathcal{X}^2(m-1), \quad \dfrac{(n-1)s_y^2}{\sigma_2^2} \sim \mathcal{X}^2(n-1)
\end{align} It can be seen
from the definition of \(F\) distribution \( F \sim F(m-1,n-1)\) .
Corollary 3: Under the above notation, set \(\sigma_1^2 = \sigma_2^2=\sigma^2\) , and note
\begin{align }
s_w^2 = \dfrac{(m-1)s_x^2+(n-1)s_y^2}{m+n-2} = \dfrac{\sum_{i=1}^m(x_i-\ bar{x})^2 + \sum_{i=1}^n(y_i-\bar{y})^2}{m+n-2}
\end{align}
then
\begin{align}
\dfrac{ (\bar{x}-\bar{y} - (\mu_1-\mu_2))}{s_w \sqrt{\dfrac{1}{m}+ \dfrac{1}{n}}} \sim t( m+n-2)
\end{align}
proves: by\(\bar{x}\sim N(\mu_1, \sigma^2 / m)\) , \(\bar{y}\sim N(\mu_2, \sigma^2 / n)\) , \( \bar{x}\) is independent of \(\bar{y}\) , so
\begin{align}
\bar{x}-\bar{y} \sim N \left( \mu_1-\mu_2, \ left( \dfrac{1}{m}+\dfrac{1}{n} \right) \sigma^2 \right)
\end{align}
so
\begin{align}
\dfrac{(\bar{x}- \bar{y} - (\mu_1-\mu_2))}{\sigma \sqrt{\dfrac{1}{m}+ \dfrac{1}{n}}} \sim N(0,1)
\end {align}
is known from the above theorem, \(\dfrac{(m-1)s_x^2}{\sigma^2}\sim \mathcal{X}^2(m-1)\) , \(\dfrac{ (n-1)s_y^2}{\sigma^2}\sim \mathcal{X}^2(n-1)\) , and they are independent of each other, then by additivity
\begin{align}
\dfrac{(m+n-2)s_w^2}{\sigma^2} = \dfrac{(m-1)s_x^2+(n-1)s_y^2}{\sigma^2} \sim \mathcal{X}^2(m+n-2)
\end{align}
Since \(\bar{x}-\bar{y}\) and \(s_w^2\) are independent of each other, according to \(t \) The definition of distribution can be concluded.