# SE of weighted mean

$X$ is a random variable with unknown distribution. A number of experiments are conducted to estimate $X$. Each experiment has a different reliability measure in estimating $X$. These $n$ experiments resulted in following sample set $\{x_1, x_2, x_3, ... , x_n\}$ with corresponding non-zero weights being $\{w_1, w_2, w_3, ... , w_n\}$. The higher weight corresponds to higher reliability. Note ${\sum_{i=1}^n{w_i}}$ can be greater than $1$.

The best unbiased estimator of true value of $X$ is the weighted mean of sample,
$\hat{X} = \bar{x}_w$,
where, $\bar{x}_w = \frac{\sum_{i=1}^n{w_ix_i}}{\sum_{i=1}^n{w_i}}$

The estimator for variance of $X$ from its true mean is,
$\hat{\sigma^2} = \bar{\sigma^2}_w$,
where, $\bar{\sigma^2}_w = \frac{\sum_{i=1}^n {w_i(x_i-\bar{x}_w)^2}}{\sum_{i=1}^n{w_i}}$,

What would be the best estimate of Standard Error of the sampling distribution of $\bar{x}_w$. Would it be $\frac{\bar{\sigma}_w}{\sqrt{n}}$. If yes, can someone help derive/explain it.

• Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. – gung Dec 18 '16 at 3:26
• Also, can you clarify why some sample means are more reliable than others? Specifically, is it for some reason other than some samples have larger N? – gung Dec 18 '16 at 3:27
• No its a practical situation I am facing. With regards to reliability, this is the nature of experimental setup. However the experiments which have higher weights have larger samples. – Gerry Dec 18 '16 at 11:06
• Is the only reason for the higher weights that they have larger N? Do you know the njs for each sample? – gung Dec 18 '16 at 13:33
• I think what @gung may be getting at is the distinction between what Wikipedia calls frequency- vs. reliability-weights. Your comment's point 2 seems to hint that both effects are perhaps present, and that you are trying to compute an expected value of some "economic function" based on "compressed samples"? (See also here for the case where the $x_i$ are not i.i.d.) – GeoMatt22 Dec 19 '16 at 1:08

Well, under i.i.d. assumptions for the $x_i$ the variance of the weighted sample mean relative to the population variance would have a factor $\left(\sum_iw_i^2\right)/\left(\sum_iw_i\right)^2$, which reduces to $1/n$ if all of the weights are equal. And there would be a comparable substitution for the "$1-1/n$" factor on the unbiased sample variance as well. See here for example.

For explanation, we have $\bar{x}=\frac{1}{W}\sum_iw_ix_i$, where $W=\sum_iw_i$.

So then using $\mathbb{V}[\,]$ to denote variance, we have \begin{align} \mathbb{V}[\bar{x}] &= \mathbb{V}\left[\tfrac{1}{W}\sum_iw_ix_i\right] \\ &=\sum_i\mathbb{V}\left[\frac{\,w_i\,}{W}\,x_i\right] \\ &=\sum_i\left[\left(\frac{\,w_i}{W}\right)^2\,\mathbb{V}[x_i]\right] \\ &=\mathbb{V}[x]\frac{\sum_iw_i^2}{W^2} \\ \end{align}