相关系数差异性检验¶
对于双侧检验，统计学假设如下：
\[
\begin{align}
H_0 & : \rho_1 = \rho_0 \\
H_1 & : \rho_1 \neq \rho_0
\end{align}
\]
对于左单侧检验，统计学假设如下：
\[
\begin{align}
H_0 & : \rho_1 \geqslant \rho_0 \\
H_1 & : \rho_1 \lt \rho_0
\end{align}
\]
对于右单侧检验，统计学假设如下：
\[
\begin{align}
H_0 & : \rho_1 \leqslant \rho_0 \\
H_1 & : \rho_1 \gt \rho_0
\end{align}
\]
样本相关系数用 \(\hat{\rho}_1\) 表示。
以下推导过程在边界条件 \(\rho_1 = \rho_0\) 下进行。
Fisher's z 转换：
\[
\zeta = \operatorname{arctanh}\rho = \frac{1}{2} \ln{\frac{1+\rho}{1-\rho}}
\]
未校正偏倚¶
在 \(H_0\) 成立时，\(\hat{\zeta}_1\) 近似服从正态分布：
\[
\hat{\zeta}_1 \sim N\left(\zeta_0,\ \frac{1}{n-3}\right)
\]
构建 \(z\) 统计量：
\[
z = \frac{\hat{\zeta}_1 - \zeta_0}{1/\sqrt{n-3}} = \left(\hat{\zeta}_1 - \zeta_0\right) \sqrt{n-3} \sim N(0, 1)
\]
在 \(H_1\) 成立时，\(\hat{\zeta}\) 近似服从正态分布：
\[
\hat{\zeta}_1 \sim N\left(\zeta_1,\ \frac{1}{n-3}\right)
\]
构建 \(z'\) 统计量：
\[
z' = \frac{\hat{\zeta}_1 - \zeta_0}{1/\sqrt{n-3}} = \left(\hat{\zeta}_1 - \zeta_0\right) \sqrt{n-3} \sim N\left(\left(\zeta_1 - \zeta_0\right)\sqrt{n-3}, 1\right)
\]
\[
\begin{aligned}
\text{Power}
= P\left(z' > z_{1-\alpha/2}\right) + P\left(z' < z_{\alpha/2}\right)
= 1 - \Phi\left(z_{1-\alpha/2} - \left(\zeta_1 - \zeta_0\right) \sqrt{n-3}\right) + \Phi\left(z_{\alpha/2} - \left(\zeta_1 - \zeta_0\right) \sqrt{n-3}\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{Power}
= P\left(z' < z_{\alpha}\right)
= \Phi\left(z_{\alpha} - \left(\zeta_1 - \zeta_0\right) \sqrt{n-3}\right)
= 1 - \Phi\left(z_{1-\alpha} + \left(\zeta_1 - \zeta_0\right) \sqrt{n-3}\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{Power}
= P\left(z' > z_{1-\alpha}\right)
= 1 - \Phi\left(z_{1-\alpha} - \left(\zeta_1 - \zeta_0\right) \sqrt{n-3}\right)
\end{aligned}
\]
对于单侧检验，可利用功效函数得到样本量的闭式解：
根据标准正态分布分位数的定义：
\[
z_{1-\alpha} \pm \left(\zeta_1 - \zeta_0\right) \sqrt{n-3} = z_{\beta}
\]
可解出：
\[
n = \frac{\left(z_{1-\alpha} + z_{1-\beta}\right)^2}{\left(\zeta_1 - \zeta_0\right)^2} + 3
\]
利用 Fisher's z 转换得：
\[
n = 4 \left( \frac{z_{1-\alpha} + z_{1-\beta}}{\ln{\frac{\left(1+\rho_1\right) \left(1-\rho_0\right)}{\left(1-\rho_1\right) \left(1+\rho_0\right)}}} \right)^2+ 3
\]
校正偏倚¶
在 未校正偏倚 的基础上添加校正项 \(\frac{\rho}{2(n-1)}\)。
在 \(H_0\) 成立时，\(\hat{\zeta}_1\) 近似服从正态分布：
\[
\hat{\zeta}_1 \sim N\left(\zeta_0 + \frac{\rho_0}{2(n-1)}, \frac{1}{n-3}\right)
\]
构建 \(z\) 统计量：
\[
z = \frac{\hat{\zeta}_1 - \zeta_0 - \frac{\rho_0}{2(n-1)}}{1/\sqrt{n-3}} = \left(\hat{\zeta}_1 - \zeta_0 - \frac{\rho_0}{2(n-1)}\right) \sqrt{n-3} \sim N(0, 1)
\]
在 \(H_1\) 成立时，\(\hat{\zeta}_1\) 近似服从正态分布：
\[
\hat{\zeta}_1 \sim N\left(\zeta_1 + \frac{\rho_1}{2(n-1)},\ \frac{1}{n-3}\right)
\]
构建 \(z'\) 统计量：
\[
z' = \frac{\hat{\zeta}_1 - \zeta_0 - \frac{\rho_0}{2(n-1)}}{1/\sqrt{n-3}}
= \left(\hat{\zeta}_1 - \zeta_0 - \frac{\rho_0}{2(n-1)}\right) \sqrt{n-3}
\sim N\left(\left(\zeta_1 - \zeta_0 + \frac{\rho_1 - \rho_0}{2(n-1)}\right) \sqrt{n-3}, 1\right)
\]
\[
\begin{aligned}
\text{Power}
& = P\left(z' > z_{1-\alpha/2}\right) + P\left(z' < z_{\alpha/2}\right) \\
& = 1 - \Phi\left(z_{1-\alpha/2} - \left(\zeta_1 - \zeta_0 + \frac{\rho_1 - \rho_0}{2(n-1)}\right) \sqrt{n-3}\right)
+ \Phi\left(z_{\alpha/2} - \left(\zeta_1 - \zeta_0 + \frac{\rho_1 - \rho_0}{2(n-1)}\right) \sqrt{n-3}\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{Power}
& = P\left(z' < z_{\alpha}\right) \\
& = \Phi\left(z_{\alpha} - \left(\zeta_1 - \zeta_0 + \frac{\rho_1 - \rho_0}{2(n-1)}\right) \sqrt{n-3}\right) \\
& = 1 - \Phi\left(z_{1-\alpha} + \left(\zeta_1 - \zeta_0 + \frac{\rho_1 - \rho_0}{2(n-1)}\right) \sqrt{n-3}\right)
\end{aligned}
\]
\[
\begin{aligned}
\text{Power}
& = P\left(z' > z_{1-\alpha}\right) \\
& = 1 - \Phi\left(z_{1-\alpha} - \left(\zeta_1 - \zeta_0 + \frac{\rho_1 - \rho_0}{2(n-1)}\right) \sqrt{n-3}\right)
\end{aligned}
\]