给定样本平均值，样本中位数的期望值

让$Y$表示中值，并让$\bar{X}$表示平均值，大小的随机样本的$n=2k+1$从分发即$N(\mu,\sigma^2)$。我该如何计算$E(Y|\bar{X}=\bar{x})$？

直观地说，因为态假设的，是有意义的要求是$E(Y|\bar{X}=\bar{x})=\bar{x}$的确是正确的答案。可以严格显示吗？

我最初的想法是使用条件正态分布来解决此问题，这通常是已知的结果。那里的问题是，由于我不知道期望值，因此也不知道中位数的方差，因此我将不得不使用$k+1$阶统计量来计算那些值。但这非常复杂，除非绝对必要，否则我不愿去那里。

让X表示原始样品和Ž随机矢量的条目Ž ķ = X ķ - ˉ X。然后Z以法线为中心（但其项不是独立的，从它们的总和为零（很有可能）可以看出）。作为线性功能的X，载体（Ž ，ˉ X）是正常的，因此它的协方差矩阵就足够的计算表明ž是独立的ˉ X。$X$$Z$$Z_k=X_k-\bar X$$Z$$X$$(Z,\bar X)$$Z$$\bar X$

转到ÿ，人们可以看出Ŷ = ˉ X + Ť其中Ť是的中值Ž。特别是，Ť取决于Ž只因此Ť是独立的ˉ X，和分布Ž是对称的，因此Ť居中。$Y$$Y=\bar X+T$$T$$Z$$T$$Z$$T$$\bar X$$Z$$T$

最后，

$$E(Y\mid\bar X)=\bar X+E(T\mid\bar X)=\bar X+E(T)=\bar X.$$

样本中位数是一个顺序统计量，并且具有非正态分布，因此样本中位数和样本均值（具有正态分布）的联合有限样本分布不会是双变量正态。借助近似，渐近地成立以下条件（请参阅此处的答案）：

$$\sqrt n\Big [\left (\begin{matrix} \bar X_n \\ Y_n \end{matrix}\right) - \left (\begin{matrix} \mu \\ \mathbb v \end{matrix}\right)\Big ] \rightarrow_{\mathbf L}\; N\Big [\left (\begin{matrix} 0 \\ 0 \end{matrix}\right) , \Sigma \Big]$$

与

$$\Sigma = \left (\begin{matrix} \sigma^2 & E\left( |X-\mathbb v|\right)\left[2f(\mathbb v)\right]^{-1} \\ E\left(|X-\mathbb v|\right)\left[2f(\mathbb v)\right]^{-1} & \left[2f(\mathbb v)\right]^{-2} \end{matrix}\right)$$

where $\bar X_n$ is the sample mean and $\mu$ the population mean, $Y_n$ is the sample median and $\mathbb v$ the population median, $f()$ is the probability density of the random variables involved and $\sigma^2$ is the variance.

So approximately for large samples, their joint distribution is bivariate normal, so we have that

$$E(Y_n \mid \bar X_n=\bar x) = \mathbb v + \rho\frac {\sigma_{\mathbb v}}{\sigma_{\bar X}}(\bar x -\mu)$$

where $\rho$ is the correlation coefficient.

Manipulating the asymptotic distribution to become the approximate large-sample joint distribution of sample mean and sample median (and not of the standardized quantities), we have

$$\rho = \frac {\frac 1nE\left(|X-\mathbb v|\right)\left[2f(\mathbb v)\right]^{-1}}{\frac 1n \sigma \left[2f(\mathbb v)\right]^{-1}} = \frac {E\left(|X-\mathbb v|\right)}{\sigma }$$

So

$$E(Y_n \mid \bar X_n=\bar x) = \mathbb v + \frac {E\left(|X-\mathbb v|\right)}{\sigma }\frac {\left[2f(\mathbb v)\right]^{-1}}{\sigma}(\bar x -\mu)$$

We have that $2f(\mathbb v) = 2/\sigma\sqrt{2\pi}$ due to the symmetry of the normal density so we arrive at

$$E(Y_n \mid \bar X_n=\bar x) = \mathbb v + \sqrt{\frac {\pi}{2}}E\left(\left|\frac {X-\mu}{\sigma}\right|\right)(\bar x -\mu)$$

where we have used $\mathbb v = \mu$. Now the standardized variable is a standard normal, so its absolute value is a half-normal distribution with expected value equal to $\sqrt{2/\pi}$ (since the underlying variance is unity). So

$$E(Y_n \mid \bar X_n=\bar x) = \mathbb v + \sqrt{\frac {\pi}{2}}\sqrt{\frac {2}{\pi}}(\bar x -\mu) = \mathbb v + \bar x -\mu = \bar x$$

The answer is $\bar{x}$.

Let $x = (x_1, x_2, \ldots, x_n)$ have a multivariate distribution $F$ for which all the marginals are symmetric about a common value $\mu$. (It does not matter whether they are independent or even are identically distributed.) Define $\bar{x}$ to be the arithmetic mean of the $x_i,$ $\bar{x} = (x_1+x_2+\cdots+x_n)/n$ and write $x-\bar{x} = (x_1-\bar{x}, x_2-\bar{x}, \ldots, x_n-\bar{x})$ for the vector of residuals. The symmetry assumption on $F$ implies the distribution of $x - \bar{x}$ is symmetric about $0$; that is, when $E\subset\mathbb{R}^n$ is any event,

$${\Pr}_F(x - \bar{x}\in E) = {\Pr}_F(x - \bar{x}\in -E).$$

Applying the generalized result at /stats//a/83887 shows that the median of $x-\bar{x}$ has a symmetric distribution about $0$. Assuming its expectation exists (which is certainly the case when the marginal distributions of the $x_i$ are Normal), that expectation has to be $0$ (because the symmetry implies it equals its own negative).

Now since subtracting the same value $\bar{x}$ from each of a set of values does not change their order, $Y$ (the median of the $x_i$) equals $\bar{x}$ plus the median of $x-\bar{x}$. Consequently its expectation conditional on $\bar{x}$ equals the expectation of $x-\bar{x}$ conditional on $\bar{x}$, plus $E(\bar{x}\ |\ \bar{x})$. The latter obviously is $\bar{x}$ whereas the former is $0$ because the unconditional expectation is $0$. Their sum is $\bar{x},$ QED.

This is simpler than the above answers make it. The sample mean is a complete and sufficient statistic (when the variance is known, but our results do not depend on the variance, hence will be valid also in the situation when the variance is unknown). Then the Rao-Blackwell together with the Lehmann-Scheffe theorems (see wikipedia ...) will imply that the conditional expectation of the median, given the arithmetic mean, is the unique minimum variance unbiased estimator of the expectation $\mu$. But we know that is the arithmetic mean, hence the result follows.

We did also use that the median is an unbiased estimator, which follows from symmetry.