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# 统计代写|The Correlation抽样理论代考

## 统计代写

• Let $Z_{1}$ and $Z_{2}$ be two independent standard normal random variables. Let $\sigma_{1}>$ $0, \sigma_{2}>0,-1<\rho<1, \mu_{1}$, and $\mu_{2}$ be real numbers. Define
$$X_{1}=\sigma_{1} Z_{1}+\mu_{1} \text { and } X_{2}=\sigma_{2} \rho Z_{1}+\sigma_{2} \sqrt{1-\rho^{2}} Z_{2}+\mu_{2}$$
Then, $\left(X_{1}, X_{2}\right)$ is said to have a bivariate normal distribution with parameters $\mu_{1}, \mu_{2}, \sigma_{1}, \sigma_{2}$, and $\rho$.
• Assume $\left(X_{1}, X_{2}\right)$ is a bivariate normal vector with parameters $\mu_{1}, \mu_{2}, \sigma_{1}, \sigma_{2}$, and $\rho$. Then $X_{1}$ is normally distributed with mean $\mu_{1}$ and variance $\sigma_{1}^{2}$ and $X_{2}$ is normally distributed with mean $\mu_{2}$ and variance $\sigma_{2}^{2}$.
It is easy to see that $X_{1}$ is normally distributed as a linear transformation of a normal random variable $Z_{1}$. The fact that $X_{2}$ is normally distributed comes from the following property of the normal distribution. A linear combination of two independent normal variables is also normal. This will be proved in the moment random variables $Z_{1}$ and $Z_{2}, X_{2}$ is also normally distributed.
We will show the following properties of a bivariate normal:
• Consider a bivariate random vector $\left(X_{1}, X_{2}\right)$ as defined above. Then, $E\left(X_{1}\right)=$ $\mu_{1}, E\left(X_{2}\right)=\mu_{2}, \operatorname{Var}\left(X_{1}\right)=\sigma_{1}^{2}, \operatorname{Var}\left(X_{2}\right)=\sigma_{2}^{2}$, and the correlation of $\left(X_{1}, X_{2}\right)$ is $\rho$.
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We now perform these computations. We will use that $Z_{1}$ and $Z_{2}$ are independent, $E\left(Z_{1}\right)=E\left(Z_{2}\right)=0$, and $\operatorname{Var}\left(Z_{1}\right)=\operatorname{Var}\left(Z_{2}\right)=1$. Note that these
(C) Springer Nature Switzerland AG 2022 R. B. Schinazi, Probability with Statistical Applications,
Using that $E\left(Z_{1}\right)=E\left(Z_{2}\right)=0$ and the linearity of the expectation it is easy to that $E\left(X_{1}\right)=\mu_{1}$ and $E\left(X_{2}\right)=\mu_{2}$. This is left to the reader. Using that the variance is shift invariant and is quadratic, $$\operatorname{Var}\left(X_{1}\right)=\operatorname{Var}\left(\sigma_{1} Z_{1}+\mu_{1}\right)$$ $$=\operatorname{Var}\left(\sigma_{1} Z_{1}\right)$$ $$=\sigma_{1}^{2} \operatorname{Var}\left(Z_{1}\right)$$ $$=\sigma_{1}^{2}$$
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computations hold for any distributions for $Z_{1}$ and $Z_{2}$. The fact that $Z_{1}$ and $Z_{2}$ are normally distributed plays no role at this point.

Using that $E\left(Z_{1}\right)=E\left(Z_{2}\right)=0$ and the linearity of the expectation it is easy to see that $E\left(X_{1}\right)=\mu_{1}$ and $E\left(X_{2}\right)=\mu_{2}$. This is left to the reader.
Using that the variance is shift invariant and is quadratic,
We now turn to $\operatorname{Var}\left(X_{2}\right)$, using the independence of $Z_{1}$ and $Z_{2}$,
We now compute the covariance of $X_{1}$ and $X_{2}$. Recall that the covariance is shift \begin{aligned} \operatorname{Cov}\left(X_{1}, X_{2}\right) &=\operatorname{Cov}\left(\sigma_{1} Z_{1}+\mu_{1}, \sigma_{2} \rho Z_{1}+\sigma_{2} \sqrt{1-\rho^{2}} Z_{2}+\mu_{2}\right) \ &=\operatorname{Cov}\left(\sigma_{1} Z_{1}, \sigma_{2} \rho Z_{1}+\sigma_{2} \sqrt{1-\rho^{2}} Z_{2}\right) \end{aligned} invariant and linear in both components, hence
$$=\sigma_{1} \sigma_{2} \rho \operatorname{Cov}\left(Z_{1}, Z_{1}\right)+\sigma_{1} \sigma_{2} \sqrt{1-\rho^{2}} \operatorname{Cov}\left(Z_{1}, Z_{2}\right)$$
Using now that $\operatorname{Cov}\left(Z_{1}, Z_{1}\right)=\operatorname{Var}\left(Z_{1}\right)=1$ and that $\operatorname{Cov}\left(Z_{1}, Z_{2}\right)=0$ we get
$$\operatorname{Cov}\left(X_{1}, X_{2}\right)=\sigma_{1} \sigma_{2} \rho$$
By definition

• 令 $Z_{1}$ 和 $Z_{2}$ 是两个独立的标准正态随机变量。令 $\sigma_{1}>$ $0, \sigma_{2}>0,-1<\rho<1, \mu_{1}$ 和 $\mu_{2}$ 为实数。定义
$$X_{1}=\sigma_{1} Z_{1}+\mu_{1} \text { 和 } X_{2}=\sigma_{2} \rho Z_{1}+\sigma_{2} \sqrt{ 1-\rho^{2}} Z_{2}+\mu_{2}$$
然后，$\left(X_{1}, X_{2}\right)$ 被称为具有参数 $\mu_{1}, \mu_{2}, \sigma_{1}, \sigma_ 的二元正态分布{2}$ 和 $\rho$。
• 假设 $\left(X_{1}, X_{2}\right)$ 是具有参数 $\mu_{1}, \mu_{2}, \sigma_{1}, \sigma_{2} 的二元法线向量$ 和 $\rho$。那么 $X_{1}$ 正态分布，均值 $\mu_{1}$ 和方差 $\sigma_{1}^{2}$ 并且 $X_{2}$ 正态分布，均值 $\mu_{2}$ 和方差 $\sigma_{2}^{2}$。
很容易看出，$X_{1}$ 是作为正态随机变量 $Z_{1}$ 的线性变换的正态分布。 $X_{2}$ 是正态分布的事实来自正态分布的以下性质。两个独立正态变量的线性组合也是正态的。这将在随机变量 $Z_{1}$ 和 $Z_{2} 的时刻得到证明，X_{2}$ 也是正态分布的。
我们将展示二元正态的以下属性：