mathematical-statisticsprobabilityrandom variable
In the Elements of Statistical Learning book at page 588 the author states that the average of $B$ i.i.d. random variables, each with variance $\sigma^2$, has variance $\frac{1}{B} \sigma^2$.
If the variables are simply i.d. (identically distributed, but not
necessarily independent) with positive pairwise correlation $\rho$, the variance of the average is
$$\rho \sigma^2 + \frac{1 – \rho}{B} \sigma^2.$$
I don't understand the calculation of the second case.
The median is the central order statistic when the number of observations is odd. If $n$ is even then the median is either an order statistic, or the mean of 2 order statistics (or something else) depending on which definition of median you use.
So the exact distribution of the median can be worked out based on the distribution of order statistics. For odd $n$ where all the $x$'s are iid from a pdf $f$ with cumulative distribution $F$ the distribution of the median is:
$\binom{n-1}{(n-1)/2} F(x)^{\frac{n-1}2} f(x) (1-F(x))^{\frac{n-1}2}$
You can google "distribution of order statistics" to get more details and derivation.
For the normal we don't have a closed form solution for $F(x)$, but there are computational tools that can help estimate the above (see the distr package for R for one possibility).
If your main goal is just an estimate of the variance of the median, then a simpler approach is just to simulate a bunch of datasets and compute the variance of their medians (and the variance of their means for comparison).
The Wikipedia article on "Median" also has information that may be of interest.
There are different way to see this.
1
The top side (y positive) is a (point symmetric) mirror image of the bottom side (y negative). So you can express $P(\frac{x}{y}|y<0)$ by $P(\frac{x}{y}|y>0)$ and similarly $P(\frac{x}{|y|}|y<0)$ by $P(\frac{x}{|y|}|y>0)$
2
thus effectively you flip the bottom half from $\frac{x}{y}=b$ to $\frac{x}{y}=-b$ but this half is symmetric in $P(\frac{x}{y}|y<0) = P(-\frac{x}{y}|y<0)$
3
a conversion to cyclic coordinates may also work. $P(\frac{x}{y}=a) \propto 2 P(\theta=tan^{-1}(a))$ and also $P(\frac{x}{|y|}=a) \propto 2 P(\theta=tan^{-1}(a))$
image: 1000 points x,y distributed i.i.d. N(0,1)
Best Answer
Let $X_1, \dots, X_B$ be the corresponding random variables, and let $$\bar{X}_B = \dfrac{1}{B}\sum_{i=1}^{B}X_i$$ be their average.
Then
$$\text{Var}(\bar{X}_B) = \dfrac{1}{B^2}\text{Var}\left(\sum_{i=1}^{B}X_i\right) = \dfrac{1}{B^2}\sum_{i=1}^{B}\sum_{j=1}^{B}\text{Cov}(X_i, X_j)$$ Suppose, in the above summation, that $i = j$. Then $\text{Cov}(X_i, X_j) = \sigma^2$. Exactly $B$ of these occur.
Suppose, in the above summation, that $i \neq j$. Then $\text{Cov}(X_i, X_j) = \rho\sigma^2$ since the variances are identical. There are $B^2 - B = B(B-1)$ of these occurrences. (Notice that there are $B * B = B^2$ total terms in the summmation, so $B^2 - B$ is the number of terms that aren't equal to $\sigma^2$, as above.)
Hence, $$\sum_{i=1}^{B}\sum_{j=1}^{B}\text{Cov}(X_i, X_j) = B\sigma^2+B(B-1)\rho\sigma^2$$ from which we obtain $$\text{Var}(\bar{X}_B) = \dfrac{1}{B^2}\left(B\sigma^2+B(B-1)\rho\sigma^2\right) = \dfrac{\sigma^2}{B}+\dfrac{B-1}{B}\rho\sigma^2 = \dfrac{\sigma^2}{B}+\rho\sigma^2-\dfrac{1}{B}\rho\sigma^2$$ or $$\rho\sigma^2 +\dfrac{\sigma^2}{B}(1-\rho)$$ as desired.