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Why no variance term in Bayesian logistic regression?


Bayesian logit model - intuitive explanation?Are logistic regression coefficient estimates biased when the predictor has large variance?logistic regression with slackClosed form for the variance of a sum of two estimates in logistic regression?Evaluate posterior predictive distribution in Bayesian linear regressionClassical vs Bayesian logistic regression assumptionsCase Control Sampling in Logistic RegressionFit logistic regression with linear constraints on coefficients in RIs Bayesian Ridge Regression another name of Bayesian Linear Regression?Bayesian Inference for More Than Linear RegressionEconometrics: What are the assumptions of logistic regression for causal inference?













2












$begingroup$


I've read here that




... (Bayesian linear regression) is most similar to Bayesian inference
in logistic regression, but in some ways logistic regression is even
simpler, because there is no variance term to estimate, only the
regression parameters.




Why is it the case, why no variance term in Bayesian logistic regression?










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    I've read here that




    ... (Bayesian linear regression) is most similar to Bayesian inference
    in logistic regression, but in some ways logistic regression is even
    simpler, because there is no variance term to estimate, only the
    regression parameters.




    Why is it the case, why no variance term in Bayesian logistic regression?










    share|cite|improve this question









    $endgroup$














      2












      2








      2


      1



      $begingroup$


      I've read here that




      ... (Bayesian linear regression) is most similar to Bayesian inference
      in logistic regression, but in some ways logistic regression is even
      simpler, because there is no variance term to estimate, only the
      regression parameters.




      Why is it the case, why no variance term in Bayesian logistic regression?










      share|cite|improve this question









      $endgroup$




      I've read here that




      ... (Bayesian linear regression) is most similar to Bayesian inference
      in logistic regression, but in some ways logistic regression is even
      simpler, because there is no variance term to estimate, only the
      regression parameters.




      Why is it the case, why no variance term in Bayesian logistic regression?







      logistic bayesian variance






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      PatrickPatrick

      1396




      1396




















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            @patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
            $endgroup$
            – seanv507
            57 mins ago










          • $begingroup$
            @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
            $endgroup$
            – Patrick
            23 mins ago







          • 1




            $begingroup$
            @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
            $endgroup$
            – Tim
            21 mins ago











          Your Answer





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          1 Answer
          1






          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            @patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
            $endgroup$
            – seanv507
            57 mins ago










          • $begingroup$
            @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
            $endgroup$
            – Patrick
            23 mins ago







          • 1




            $begingroup$
            @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
            $endgroup$
            – Tim
            21 mins ago















          5












          $begingroup$

          Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            @patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
            $endgroup$
            – seanv507
            57 mins ago










          • $begingroup$
            @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
            $endgroup$
            – Patrick
            23 mins ago







          • 1




            $begingroup$
            @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
            $endgroup$
            – Tim
            21 mins ago













          5












          5








          5





          $begingroup$

          Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.






          share|cite|improve this answer









          $endgroup$



          Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          TimTim

          59.6k9131224




          59.6k9131224











          • $begingroup$
            @patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
            $endgroup$
            – seanv507
            57 mins ago










          • $begingroup$
            @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
            $endgroup$
            – Patrick
            23 mins ago







          • 1




            $begingroup$
            @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
            $endgroup$
            – Tim
            21 mins ago
















          • $begingroup$
            @patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
            $endgroup$
            – seanv507
            57 mins ago










          • $begingroup$
            @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
            $endgroup$
            – Patrick
            23 mins ago







          • 1




            $begingroup$
            @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
            $endgroup$
            – Tim
            21 mins ago















          $begingroup$
          @patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
          $endgroup$
          – seanv507
          57 mins ago




          $begingroup$
          @patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
          $endgroup$
          – seanv507
          57 mins ago












          $begingroup$
          @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
          $endgroup$
          – Patrick
          23 mins ago





          $begingroup$
          @seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
          $endgroup$
          – Patrick
          23 mins ago





          1




          1




          $begingroup$
          @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
          $endgroup$
          – Tim
          21 mins ago




          $begingroup$
          @Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
          $endgroup$
          – Tim
          21 mins ago

















          draft saved

          draft discarded
















































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