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  • Negative delta when using psacalc (Oster, 2016)

    Dear all,

    I am trying to assess the omitted variable bias using the methodology presented by Oster (2016) for a coefficient for oil_ever which is significantly positive in the controlled regression.

    I typed the following command in stata :
    psacalc delta oil_ever
    And stata returned this:

    ---- Bound Estimate ----
    -------------+----------------------------------------------------------------
    delta | -2.97194
    -------------+----------------------------------------------------------------

    ---- Inputs from Regressions ----
    | Coeff. R-Squared
    -------------+----------------------------------------------------------------
    Uncontrolled | 0.40441 0.003
    Controlled | 1.17795 0.501
    -------------+----------------------------------------------------------------

    ---- Other Inputs ----
    -------------+----------------------------------------------------------------
    R_max | 1.000
    Beta | 0.000000
    Unr. Controls|
    -------------+----------------------------------------------------------------

    I am confused about the negative sign of delta. I understand that this means selection on unobservables should be of the opposite sign that selection on the observables for the effect to be explained away by the bias as explained here Unobservable Selection(Oster 2016) Using psacalc by one of the package's author. However what I am confused about is the following : when \delta is positive, the selection on unobservables should be at least \delta*(selection on observables) for the effect to be 0. When obtaining a negative \delta : for the omitted variable bias to explain away the effect of the treatment variable, does the \delta need to be bigger -2.97 (meaning if selection on unobservables is -1*(selection on observables), the conclusion would be that the effect found is driven by the bias) or does it have to be smaller than -2.97 i.e. bigger in absolute value but negative (in that case, the selection bias on unobservables needs to be pretty important relative to the one on observables and negative, which is quite unlikely, so I would consider my result robust) ?

    I hope I am explaining things clearly.
    Many thanks in advance for your help.

    Anne
    Last edited by Anne Fre; 02 Jun 2022, 02:30.

  • #2
    The larger delta in absolute value, the stronger the endogeneity. However, notice that the strength of the selection bias, as expressed by delta, is not symmetric around zero. Due to its nonlinearity, a delta equal to +1 is not fully comparable to a delta equal to -1.

    We have a brief discussion and further references in the following paper, where we suggest an alternative approach for robustness analysis:
    Last edited by Sebastian Kripfganz; 02 Jun 2022, 05:06.
    https://www.kripfganz.de/stata/

    Comment


    • #3
      Thank you for your answer, I will definitely look into your paper.
      am not sure I understand what you mean by "the larger delta in absolute value, the stronger the endogeneity" though. Does this mean that selection on unobservables must be very large (even if of opposite sign potentially) compared to selection on observables to explain entirely the coefficient when delta is large in absolute value ?

      Comment


      • #4
        To be honest, I always struggled with the phrase "selection on observables/unobservables". What does it mean in intuitive/practical terms that selection on unobservables is, say, twice as large as selection on observables? And related to that, why is "equal selection on observables and unobservables" a good reference point?

        I always think about it in terms of the endogeneity of the observables in the regression model, i.e. the correlation of the observed regressors with the regression error (which includes the unobservables). The larger the endogeneity, the larger the bias. The larger delta in absolute value, the larger the endogeneity of the regressors (at least in a neighborhood of 0).

        A correlation is easy to interpret. Delta gives me a headache when it comes to interpreting its value.
        Last edited by Sebastian Kripfganz; 03 Jun 2022, 05:28.
        https://www.kripfganz.de/stata/

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        • #5
          Thank you for your help ! It took me a bit of time to see it your way but I think I understand what you mean.

          Comment


          • #6
            There following might be of interest as well:
            https://www.kripfganz.de/stata/

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