Matlab加权方差

Question

我研究了如何计算样本的加权标准差或方差，并发现这篇文章参考了Gatz和Smith提出的几种方法：https://math.stackexchange.com/questions/823125/sampling-error-with-weighted-mean。

现在我试着理解Matlab在使用"std(A，w)“时是如何计算加权方差的。我使用"edit“查看了Matlab和原始代码，但未能理解如何计算加权方差的语法。有人能给我写下这个方程式，或者用几句话来描述下面代码的关键行吗？(在Matlab中键入"edit“即可找到函数的完整代码)。

% Weighted variance

否则

if ~isvector(w) || ~isreal(w) || ~isfloat(w) || ...
   (omitnan && ~all(w(~isnan(w)) >= 0)) || (~omitnan && ~all(w >= 0))
    error(message('MATLAB:var:invalidWgts'));
end

if numel(w) ~= n
    if isscalar(w)
        error(message('MATLAB:var:invalidWgts'));
    else
        error(message('MATLAB:var:invalidSizeWgts'));
    end
end

if ~omitnan

    % Normalize W, and embed it in the right number of dims.  Then
    % replicate it out along the non-working dims to match X's size.
    wresize = ones(1,max(ndims(x),dim)); wresize(dim) = n;
    w = reshape(w ./ sum(w), wresize);
    y = sum(w .* abs(x  - sum(w .* x, dim)).^2, dim); % abs guarantees a real result

else
    % Repeat vector W, such that new W has the same size as X
    sz = size(x); sz(end+1:dim) = 1;
    wresize = ones(size(sz)); wresize(dim) = sz(dim);
    wtile = sz; wtile(dim) = 1;
    w = repmat(reshape(w, wresize), wtile);

    % Count up non-NaN weights at non-NaN elements
    w(isnan(x)) = NaN;
    denom = sum(w, dim, 'omitnan'); % contains no NaN, since w >= 0

    x = x - (sum(w .* x, dim, 'omitnan') ./ denom);
    wx2 = w .* abs(x).^2;
    y = sum(wx2, dim, 'omitnan') ./ denom; % abs guarantees a real result

    % Don't omit NaNs caused by computation (not missing data)
    ind = any(isnan(wx2) & ~isnan(w), dim);
    y(ind) = NaN;

Answer 1

所以，我相信刚才提到的方程式有点不正确。

sqrt(sum((x-mean(x)).^2.*w(:))/sum(w)) % slightly off

需要根据数据的长度对数据进行加权，以匹配MATLAB的std()输出，并匹配以下基本STD方程：

sqrt(sum((x(:)-mean(x)).^2)/(length(x)-1)) % base STD equation

通过计算长度，我们得到以下信息：

sqrt(sum((x(:)-mean(x)).^2.*w(:))/sum(w)*length(x)/(length(x)-1)) % corrected

下面的示例演示了这些差异：

% define weighted std functions
std_fcn_a = @(x,w) sqrt(sum((x(:)-mean(x)).^2.*w(:))/sum(w));
std_fcn_b = @(x,w) sqrt(sum((x(:)-mean(x)).^2.*w(:))/sum(w)*length(x)/(length(x)-1));

% create data
x = reshape(magic(3),[1,9]);    % create example data
w0 = ones(size(x))/length(x);   % create uniform weighting
rng(0);                         % seed random data
w1 = rand(size(x));             % create random weights
w1 = w1/sum(w1);                % normalize weights

% test unweighted
std_0 = std(x)                  % 2.7386
std_0_v2 = std(x,w0)            % 2.5820
std_a_0 = std_fcn_a(x,w0)       % 2.5820
std_b_0 = std_fcn_b(x,w0)       % 2.7386

% test weighted
std_a_1 = std_fcn_a(x,w1)       % 2.6963
std_b_1 = std_fcn_b(x,w1)       % 2.8599

示例输出

std_0 = 2.7386
std_0_v2 = 2.5820
std_a_0 = 2.5820
std_b_0 = 2.7386
std_a_1 = 2.6963
std_b_1 = 2.8599

未加权检验结果表明，如果w是一致的，则std(x)与std(x，w)是不一样的。我最初假设它应该是，因为std(x)将有一个统一的加权，因此将是相等的。这就引出了我自己的问题和std(x，w)的有效性。