Weighted Least Square (WLS) Vs. Ordinary Least Square(OLS) Regression

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  • Опубликовано: 11 сен 2024
  • Weighted Least Square (WLS) regression models are fundamentally different from the Ordinary Least Square Regression (OLS) . WLS is used when the error terms are not homoskedastic. Ordinary least squares (OLS) assumes that there is constant variance in the errors (which is called homoscedasticity)
    The method of weighted least squares can be used when the ordinary least squares assumption of constant variance in the errors is violated (which is called heteroscedasticity)
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Комментарии • 10

  • @BrandonJoeReich
    @BrandonJoeReich 4 года назад

    Thank you for the video. At around 25:45 in the video you said "You regress this as depending variable and this as independent variable". I believe you misspoke because you called Y the independent variable but I just want to be sure. If I'm wrong then that means the OLS fitted values should be the independent variable and go on the x-axis even though you're pointing to Y in the video. Thank you!

  • @paulogorayeb
    @paulogorayeb 6 лет назад

    Excellent... thank you!!

  • @korman9872
    @korman9872 Год назад

    Tx sir

  • @inserthere6387
    @inserthere6387 5 лет назад

    thanks for the matrix notation

  • @looploop6612
    @looploop6612 5 лет назад

    how to do step 5?

  • @RikardoAHP
    @RikardoAHP 5 лет назад

    How do I calculate R-squared for the WLS model?

    • @joaojulio435
      @joaojulio435 3 года назад

      I think is the same way that you calculate for OLS, but now using the values of WLS. If you do a descritive summary of the model/fit in R you can see the R- Squared for the WLS model

  • @14IEA
    @14IEA 7 лет назад +1

    Excellent video!!, I have a question: If I have a matrix (X) 4x4, how can I obtained the matrix W of that matrix?
    thanks!

    • @tomasnobrega8087
      @tomasnobrega8087 3 года назад

      If X is a square matrix you will have problems with regression. The linear subspace defined by X is the same dimension of Y and so the linear subspace of residuals is zero. All dimensions of Y contained inside X. You don't want that. Can't really tell the answer to your question about W