统计中,μ和x̄的区别是什么?
发布网友
发布时间：2022-04-26 14:38
共2个回答
热心网友
时间：2022-06-29 07:31
看了你下一个问题, 应该是用英文学习的吧? 我用latex给你打了几段话，你看一下．pdf文件发不了．你可以下载latex，用代码（见下）生成．图画中的几处拼写及语法失误在code中改过了．图中难以更改，请谅解．
A few things to keep in mind
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\begin{document}
\noindent {\color{red}A one-sentence answer to your question is that $\mu$ and $\sigma$ are \textit{parameters} of the model, but $\bar{X}$ and $s$ are \textit{estimates} of $\mu$ and $\sigma$.}
\vspace{3mm} \noindent Below I elaborate
\vspace{3mm}\noindent In statistics, we often use the Greek letter $\mu$ to denote the \textit{expectation (expected value)} of a distribution
\begin{flalign*}
\mu = \left\{ \begin{array}{@{}ll@{}}
\sum x_ip_i & \text{if discrete} \\
\int X dp & \text{if continuous}
\end{array} \right.
\end{flalign*}
We consider $\mu$ \textbf{a property of the model}. On the other hand,
\begin{flalign*}
\bar{X} = \frac{1}{N} \sum^{N}_{i=1}X_i
\end{flalign*}
is the MLE (maximum likelihood estimate) of $\mu$, if $X_i$ are i.i.d. (independent and identically distributed) according to the same model.
\vspace{3mm} \noindent Example (Tossing a fair coin) Let $X$ denote the outcome of tossing a fair coin. Let $X$ equal to 1 if it lands on head, and 0 on tail. We know that $X\sim \text{Bernoulli}(\mu)$. Without any actual experimentation, we can dece that $\mu$ is 0.5 based on the (given) fact that the coin is fair. However, if we tell Bob nothing about the fairness of the coin but ask him to give an estimate of $\mu$, he will have to collect some real data. Suppose that the outcome of his first five tosses is like this
\begin{flalign*}
\{ H H T T H \}
\end{flalign*}
He will give an \textit{estimate} of $\mu$, which is $\bar{X} = \dfrac{1+1+0+0+1}{5}=0.6$
\vspace{3mm}\noindent Similarly, we often use the Greek letter $\sigma$ to denote the \textit{standard deviation} of a distribution
\begin{flalign*}
\sigma^2 = E[(x-\mu)^2]
\end{flalign*}
and $s$ to denote its estimate
\begin{flalign*}
s^2 = \frac{1}{N-1} \sum^N_{i=1}(X_i-\bar{X})^2
\end{flalign*}
\vspace{3mm} \noindent Example Continued: Based on what we know about Bernoulli distributions, the standard deviation of our model is $\sigma = \sqrt{\mu(1-\mu)}=0.5$. How would Bob estimate this parameter? Well, it's
\begin{flalign*}
& s^2 = \frac{(1-0.6)^2+(1-0.6)^2+(0-0.6)^2+(0-0.6)^2+(1-0.6)^2}{4} = 0.3 \\
& \Rightarrow s \approx 0.55
\end{flalign*}
\vspace{3mm} \noindent A few things to keep in mind
\begin{itemize}
\item $\bar{X}$ is not the only estimate of $\mu$, nor is $s^2$ of $\sigma^2$
\item People use $\bar{X}$ and $s^2$ a lot because they have nice asymptotic distributions (normal and chi-squared)
\end{itemize}
\end{document}
热心网友
时间：2022-06-29 07:31
μ和x罢都表示平均值（数学期望），前者常用于连续型随机变量，后者常用于离散型随机变量。
σ和s分别表示方差和均方差，即√σ=s.
