가중 평균 추정의 계산 표준 오차

그 가정 및 각 그려 IID 와 일부 분포로부터 독립의 . 엄격히 긍정적이다. 당신은 모든 관찰 아닌 ; 오히려 당신은 를 관찰합니다 $w_1,w_2,\ldots,w_n$ $x_1,x_2,...,x_n$ $w_i$ $x_i$ $w_i$ $w_i$ $x_i$ $\sum_i x_i w_i$ 합니다. 견적에 관심이 있습니다 $\operatorname{E}\left[x\right]$ from this information. Clearly the estimator

\bar{x} = \frac{\sum_{i} w_{i} x_{i}}{\sum_{i} w_{i}}

$\bar{x} = \frac{\sum_i w_i x_i}{\sum_i w_i}$ is unbiased, and can be computed given the information at hand.

How might I compute the standard error of this estimator? For the sub-case where $x_i$ takes only values 0 and 1, I naively tried 기본적으로의 변화를 무시하지만,이 샘플이 250 주위보다 작은 크기 (그리고 이것은 아마의 차이에 따라 약세를 발견.) 어쩌면 내가 안 보인다 '더 나은'표준 오류를 계산하기에 충분한 정보가 있습니다.

s e \approx \frac{\sqrt{\bar{x} (1 - \bar{x}) \sum_{i} w_{i}^{2}}}{\sum_{i} w_{i}},

$se \approx \frac{\sqrt{\bar{x}(1-\bar{x})\sum_i w_i^2}}{\sum_i w_i},$

w_{i}

$w_i$

w_{i}

$w_i$

standard-error weighted-mean

— 초라한 요리사
소스

답변:

최근에 같은 문제가 발생했습니다. 다음은 내가 찾은 것입니다.

가중치가 동일한 단순 랜덤 표본과 달리 가중치 평균 의 표준 오차에 대한 정의는 널리 인정되지 않습니다 . 요즘에는 부트 스트랩을 수행하고 평균의 경험적 분포를 구하는 것이 간단하며 표준 오차를 추정 한 것입니다.

이 추정을 수행하기 위해 공식을 사용하려면 어떻게해야합니까?

주요 참고 자료는 Donald F. Gatz와 Luther Smith가 작성한 이 논문 에서 3 가지 공식 기반 추정값을 부트 스트랩 결과와 비교합니다. 부트 스트랩 결과에 대한 가장 근사치는 Cochran (1977)에서 나옵니다.

$(SEM_w)^2={\dfrac{n}{(n-1)(\sum {P_i})^2}}[\sum (P_i X_i-\bar{P}\bar{X}_w)^2-2 \bar{X}_w \sum (P_i-\bar{P})(P_i X_i-\bar{P}\bar{X}_w)+\bar{X}^2_w \sum (P_i-\bar{P})^2]$

The following is the corresponding R code that came from this R listserve thread.

weighted.var.se <- function(x, w, na.rm=FALSE)
#  Computes the variance of a weighted mean following Cochran 1977 definition
{
  if (na.rm) { w <- w[i <- !is.na(x)]; x <- x[i] }
  n = length(w)
  xWbar = weighted.mean(x,w,na.rm=na.rm)
  wbar = mean(w)
  out = n/((n-1)*sum(w)^2)*(sum((w*x-wbar*xWbar)^2)-2*xWbar*sum((w-wbar)*(w*x-wbar*xWbar))+xWbar^2*sum((w-wbar)^2))
  return(out)
}

Hope this helps!

— Ming K
소스

This is pretty cool, but for my problem I don't even observe the

P_{i} X_{i}

$P_iX_i$ , rather I observe the sum

\sum_{i} P_{i} X_{i}

$\sum_i P_iX_i$ . My question is very weird because it involves some information asymmetry (a third party is reporting the sum, and trying to perhaps hide some information).

— shabbychef

Gosh you're right, sorry I did not fully understand the question you posed. Suppose we boil your problem down to the simplest case where all

w_{i}

$w_i$ are Bernoulli RV. Then you are essentially observing the sum of a random subset of

n

$n$ RVs. My guess is there is not a lot of information here to estimate with. So what did you end up doing for your original problem?

— Ming K

@Ming-ChihKao this cochran formula is interesting but if you build a confidence interval off this when the data is not normal there is no consistent interpretation correct? How would you handle non-normal weighted average mean confidence intervals? Weighted quantiles?

— user3022875

I think there is an error with the function. If you substitute w=rep(1, length(x)), then weighted.var.se(rnorm(50), rep(1, 50)) is about 0.014. I think the formula is missing a sum(w^2) in the numerator, since when P=1, the variance is 1/(n*(n-1)) * sum((x-xbar)^2). I can't check the cited article as it is behind a paywall, but I think that correction. Oddly enough, Wikipedia's (different) solution becomes degenerate when all weights are equal: en.wikipedia.org/wiki/….

— Max Candocia

These may work better in general: analyticalgroup.com/download/WEIGHTED_MEAN.pdf

— Max Candocia

The variance of your estimate given the $w_i$ is

\frac{\sum w_{i}^{2} V a r (X)}{(\sum w_{i})^{2}} = V a r (X) \frac{\sum w_{i}^{2}}{(\sum w_{i})^{2}} .

$\frac{\sum w_i^2 Var(X)}{(\sum w_i)^2} = Var(X) \frac{\sum w_i^2 }{(\sum w_i)^2}.$ Because your estimate is unbiased for any

w_{i}

$w_i$ , the variance of its conditional mean is zero. Hence, the variance of your estimate is

V a r (X) E (\frac{\sum w_{i}^{2}}{(\sum w_{i})^{2}})

$Var(X) \mathbb{E}\left(\frac{\sum w_i^2 }{(\sum w_i)^2}\right)$ With all the data observed, this would be easy to estimate empirically. But with only a measure of location of the

X_{i}

$X_i$ observed, and not their spread, I don't see how it's going to be possible to get an estimate of

V a r (X)

$Var(X)$ , without making rather severe assumptions.

— guest
소스

at least in the specific case where

x_{i}

$x_i$ have a Bernoulli distribution I can estimate the variance of

x

$x$ by

\bar{x} (1 - \bar{x})

$\bar{x}(1-\bar{x})$ as noted above. Even in this case, as noted in the question, I need a larger sample size than I would have expected.

— shabbychef