Covariance structure with one or more random intercepts












3














I need to choose a suitable covariance structure including one or more random intercepts:



If one factor B is nested in another factor A, one option would of course be to consider a model with random intercepts of both A and B, corresponding to (1|A) + (1|B) when using lmer in the R package lme4 or providing the argument random = ~1|A/B to the lme function in the R package nlme, but would it make any sense to consider a model with only the random intercept of B, even if B is nested in A?










share|cite|improve this question



























    3














    I need to choose a suitable covariance structure including one or more random intercepts:



    If one factor B is nested in another factor A, one option would of course be to consider a model with random intercepts of both A and B, corresponding to (1|A) + (1|B) when using lmer in the R package lme4 or providing the argument random = ~1|A/B to the lme function in the R package nlme, but would it make any sense to consider a model with only the random intercept of B, even if B is nested in A?










    share|cite|improve this question

























      3












      3








      3


      1





      I need to choose a suitable covariance structure including one or more random intercepts:



      If one factor B is nested in another factor A, one option would of course be to consider a model with random intercepts of both A and B, corresponding to (1|A) + (1|B) when using lmer in the R package lme4 or providing the argument random = ~1|A/B to the lme function in the R package nlme, but would it make any sense to consider a model with only the random intercept of B, even if B is nested in A?










      share|cite|improve this question













      I need to choose a suitable covariance structure including one or more random intercepts:



      If one factor B is nested in another factor A, one option would of course be to consider a model with random intercepts of both A and B, corresponding to (1|A) + (1|B) when using lmer in the R package lme4 or providing the argument random = ~1|A/B to the lme function in the R package nlme, but would it make any sense to consider a model with only the random intercept of B, even if B is nested in A?







      r regression mixed-model lme4-nlme nested-data






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 15 at 19:09









      user3419936

      255




      255






















          2 Answers
          2






          active

          oldest

          votes


















          3














          Let consider 2 levels in factor A and 2 individuals in each level of factor A, so totally 4 individuals. $i$ indicates level of factor A and $j$ for B.



          When the two intercepts are used, we have
          $$Y_{ij} = Xbeta + gamma_i + gamma_j + epsilon_{ij}$$
          $Var(gamma_i) = sigma_A^2, Var(gamma_j) = sigma_B^2, Var(epsilon) = sigma^2$



          Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
          left(begin{matrix}sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 + sigma_B^2 & sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 & sigma_A^2 & sigma^2+sigma_A^2 + sigma_B^2 &sigma_A^2 + sigma_B^2\ sigma_A^2 & sigma_A^2 & sigma_A^2 + sigma_B^2&sigma^2+sigma_A^2 + sigma_B^2 end{matrix}right)
          $$



          If intercept for factor B only, we have
          $$Y_{ij} = Xbeta + gamma_j + epsilon_{ij}$$



          Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
          left(begin{matrix}sigma^2 + sigma_B^2 & sigma_B^2 & 0&0\ sigma_B^2 & sigma^2+ sigma_B^2 & 0&0\ 0&0& sigma^2+ sigma_B^2 & sigma_B^2\ 0 & 0& sigma_B^2&sigma^2 + sigma_B^2 end{matrix}right)
          $$



          When two intercepts are in the model, $Y$ from the same individual has covariance $sigma_A^2+sigma_B^2$, $Y$ from different individuals but from same level of factor A has covariance $sigma_A^2$, which is smaller than covariance from the same individual. Generally this assumption is reasonable.



          When just individual level intercept is kept in the model, the covariance between the difference individual is zero, even they come from the same level of factor A. Generally, this is not what we want.






          share|cite|improve this answer































            3














            In general I would recommend including both (1|A) and (1|B) in the model - otherwise you're not fitting the maximal model, i.e. the model that includes all terms that are statistically identifiable. If the variation in A is reasonably large and you omit the (1|A) term, the model for the random effects corresponding to the remaining (1|B) term (that they are independently drawn from a zero-centred Normal distribution) might be off. Some reasons it might be reasonable to omit the (1|A) term:




            • there is something about your data that would eliminate any differences in A (not just experimental control: for example, maybe you've already subtracted the mean from the observations in each A group)

            • you get a singular fit, i.e. an estimate of zero variance among levels of A (in this case you'll get exactly the same answer if you drop this term)

            • you think there are too few levels of A to estimate the variance (in this case it would be more conservative to include A in the model as a fixed effect


            • if you are using data-driven model selection to decide what terms to include, and the model including (1|A) is not selected (e.g. the AIC is higher). (In general you need to be very careful when doing this kind of "sacrificial pseudoreplication" (sensu Hurlbert 1984) that you don't mess up your coverage/type-I error ...)


            (Note that the levels of B need to be uniquely coded - otherwise you should specify the nested syntax (1|A/B) (or (1|A) + (1|A:B)) in lme4 as well in nlme ...)






            share|cite|improve this answer





















              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "65"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: false,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f383156%2fcovariance-structure-with-one-or-more-random-intercepts%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3














              Let consider 2 levels in factor A and 2 individuals in each level of factor A, so totally 4 individuals. $i$ indicates level of factor A and $j$ for B.



              When the two intercepts are used, we have
              $$Y_{ij} = Xbeta + gamma_i + gamma_j + epsilon_{ij}$$
              $Var(gamma_i) = sigma_A^2, Var(gamma_j) = sigma_B^2, Var(epsilon) = sigma^2$



              Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
              left(begin{matrix}sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 + sigma_B^2 & sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 & sigma_A^2 & sigma^2+sigma_A^2 + sigma_B^2 &sigma_A^2 + sigma_B^2\ sigma_A^2 & sigma_A^2 & sigma_A^2 + sigma_B^2&sigma^2+sigma_A^2 + sigma_B^2 end{matrix}right)
              $$



              If intercept for factor B only, we have
              $$Y_{ij} = Xbeta + gamma_j + epsilon_{ij}$$



              Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
              left(begin{matrix}sigma^2 + sigma_B^2 & sigma_B^2 & 0&0\ sigma_B^2 & sigma^2+ sigma_B^2 & 0&0\ 0&0& sigma^2+ sigma_B^2 & sigma_B^2\ 0 & 0& sigma_B^2&sigma^2 + sigma_B^2 end{matrix}right)
              $$



              When two intercepts are in the model, $Y$ from the same individual has covariance $sigma_A^2+sigma_B^2$, $Y$ from different individuals but from same level of factor A has covariance $sigma_A^2$, which is smaller than covariance from the same individual. Generally this assumption is reasonable.



              When just individual level intercept is kept in the model, the covariance between the difference individual is zero, even they come from the same level of factor A. Generally, this is not what we want.






              share|cite|improve this answer




























                3














                Let consider 2 levels in factor A and 2 individuals in each level of factor A, so totally 4 individuals. $i$ indicates level of factor A and $j$ for B.



                When the two intercepts are used, we have
                $$Y_{ij} = Xbeta + gamma_i + gamma_j + epsilon_{ij}$$
                $Var(gamma_i) = sigma_A^2, Var(gamma_j) = sigma_B^2, Var(epsilon) = sigma^2$



                Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
                left(begin{matrix}sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 + sigma_B^2 & sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 & sigma_A^2 & sigma^2+sigma_A^2 + sigma_B^2 &sigma_A^2 + sigma_B^2\ sigma_A^2 & sigma_A^2 & sigma_A^2 + sigma_B^2&sigma^2+sigma_A^2 + sigma_B^2 end{matrix}right)
                $$



                If intercept for factor B only, we have
                $$Y_{ij} = Xbeta + gamma_j + epsilon_{ij}$$



                Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
                left(begin{matrix}sigma^2 + sigma_B^2 & sigma_B^2 & 0&0\ sigma_B^2 & sigma^2+ sigma_B^2 & 0&0\ 0&0& sigma^2+ sigma_B^2 & sigma_B^2\ 0 & 0& sigma_B^2&sigma^2 + sigma_B^2 end{matrix}right)
                $$



                When two intercepts are in the model, $Y$ from the same individual has covariance $sigma_A^2+sigma_B^2$, $Y$ from different individuals but from same level of factor A has covariance $sigma_A^2$, which is smaller than covariance from the same individual. Generally this assumption is reasonable.



                When just individual level intercept is kept in the model, the covariance between the difference individual is zero, even they come from the same level of factor A. Generally, this is not what we want.






                share|cite|improve this answer


























                  3












                  3








                  3






                  Let consider 2 levels in factor A and 2 individuals in each level of factor A, so totally 4 individuals. $i$ indicates level of factor A and $j$ for B.



                  When the two intercepts are used, we have
                  $$Y_{ij} = Xbeta + gamma_i + gamma_j + epsilon_{ij}$$
                  $Var(gamma_i) = sigma_A^2, Var(gamma_j) = sigma_B^2, Var(epsilon) = sigma^2$



                  Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
                  left(begin{matrix}sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 + sigma_B^2 & sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 & sigma_A^2 & sigma^2+sigma_A^2 + sigma_B^2 &sigma_A^2 + sigma_B^2\ sigma_A^2 & sigma_A^2 & sigma_A^2 + sigma_B^2&sigma^2+sigma_A^2 + sigma_B^2 end{matrix}right)
                  $$



                  If intercept for factor B only, we have
                  $$Y_{ij} = Xbeta + gamma_j + epsilon_{ij}$$



                  Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
                  left(begin{matrix}sigma^2 + sigma_B^2 & sigma_B^2 & 0&0\ sigma_B^2 & sigma^2+ sigma_B^2 & 0&0\ 0&0& sigma^2+ sigma_B^2 & sigma_B^2\ 0 & 0& sigma_B^2&sigma^2 + sigma_B^2 end{matrix}right)
                  $$



                  When two intercepts are in the model, $Y$ from the same individual has covariance $sigma_A^2+sigma_B^2$, $Y$ from different individuals but from same level of factor A has covariance $sigma_A^2$, which is smaller than covariance from the same individual. Generally this assumption is reasonable.



                  When just individual level intercept is kept in the model, the covariance between the difference individual is zero, even they come from the same level of factor A. Generally, this is not what we want.






                  share|cite|improve this answer














                  Let consider 2 levels in factor A and 2 individuals in each level of factor A, so totally 4 individuals. $i$ indicates level of factor A and $j$ for B.



                  When the two intercepts are used, we have
                  $$Y_{ij} = Xbeta + gamma_i + gamma_j + epsilon_{ij}$$
                  $Var(gamma_i) = sigma_A^2, Var(gamma_j) = sigma_B^2, Var(epsilon) = sigma^2$



                  Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
                  left(begin{matrix}sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 + sigma_B^2 & sigma^2+sigma_A^2 + sigma_B^2 & sigma_A^2&sigma_A^2 \ sigma_A^2 & sigma_A^2 & sigma^2+sigma_A^2 + sigma_B^2 &sigma_A^2 + sigma_B^2\ sigma_A^2 & sigma_A^2 & sigma_A^2 + sigma_B^2&sigma^2+sigma_A^2 + sigma_B^2 end{matrix}right)
                  $$



                  If intercept for factor B only, we have
                  $$Y_{ij} = Xbeta + gamma_j + epsilon_{ij}$$



                  Then $$Varleft(begin{matrix}Y_{11}\ Y_{12}\Y_{21}\Y_{22}end{matrix}right)=
                  left(begin{matrix}sigma^2 + sigma_B^2 & sigma_B^2 & 0&0\ sigma_B^2 & sigma^2+ sigma_B^2 & 0&0\ 0&0& sigma^2+ sigma_B^2 & sigma_B^2\ 0 & 0& sigma_B^2&sigma^2 + sigma_B^2 end{matrix}right)
                  $$



                  When two intercepts are in the model, $Y$ from the same individual has covariance $sigma_A^2+sigma_B^2$, $Y$ from different individuals but from same level of factor A has covariance $sigma_A^2$, which is smaller than covariance from the same individual. Generally this assumption is reasonable.



                  When just individual level intercept is kept in the model, the covariance between the difference individual is zero, even they come from the same level of factor A. Generally, this is not what we want.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 15 at 20:31

























                  answered Dec 15 at 19:47









                  user158565

                  5,2551318




                  5,2551318

























                      3














                      In general I would recommend including both (1|A) and (1|B) in the model - otherwise you're not fitting the maximal model, i.e. the model that includes all terms that are statistically identifiable. If the variation in A is reasonably large and you omit the (1|A) term, the model for the random effects corresponding to the remaining (1|B) term (that they are independently drawn from a zero-centred Normal distribution) might be off. Some reasons it might be reasonable to omit the (1|A) term:




                      • there is something about your data that would eliminate any differences in A (not just experimental control: for example, maybe you've already subtracted the mean from the observations in each A group)

                      • you get a singular fit, i.e. an estimate of zero variance among levels of A (in this case you'll get exactly the same answer if you drop this term)

                      • you think there are too few levels of A to estimate the variance (in this case it would be more conservative to include A in the model as a fixed effect


                      • if you are using data-driven model selection to decide what terms to include, and the model including (1|A) is not selected (e.g. the AIC is higher). (In general you need to be very careful when doing this kind of "sacrificial pseudoreplication" (sensu Hurlbert 1984) that you don't mess up your coverage/type-I error ...)


                      (Note that the levels of B need to be uniquely coded - otherwise you should specify the nested syntax (1|A/B) (or (1|A) + (1|A:B)) in lme4 as well in nlme ...)






                      share|cite|improve this answer


























                        3














                        In general I would recommend including both (1|A) and (1|B) in the model - otherwise you're not fitting the maximal model, i.e. the model that includes all terms that are statistically identifiable. If the variation in A is reasonably large and you omit the (1|A) term, the model for the random effects corresponding to the remaining (1|B) term (that they are independently drawn from a zero-centred Normal distribution) might be off. Some reasons it might be reasonable to omit the (1|A) term:




                        • there is something about your data that would eliminate any differences in A (not just experimental control: for example, maybe you've already subtracted the mean from the observations in each A group)

                        • you get a singular fit, i.e. an estimate of zero variance among levels of A (in this case you'll get exactly the same answer if you drop this term)

                        • you think there are too few levels of A to estimate the variance (in this case it would be more conservative to include A in the model as a fixed effect


                        • if you are using data-driven model selection to decide what terms to include, and the model including (1|A) is not selected (e.g. the AIC is higher). (In general you need to be very careful when doing this kind of "sacrificial pseudoreplication" (sensu Hurlbert 1984) that you don't mess up your coverage/type-I error ...)


                        (Note that the levels of B need to be uniquely coded - otherwise you should specify the nested syntax (1|A/B) (or (1|A) + (1|A:B)) in lme4 as well in nlme ...)






                        share|cite|improve this answer
























                          3












                          3








                          3






                          In general I would recommend including both (1|A) and (1|B) in the model - otherwise you're not fitting the maximal model, i.e. the model that includes all terms that are statistically identifiable. If the variation in A is reasonably large and you omit the (1|A) term, the model for the random effects corresponding to the remaining (1|B) term (that they are independently drawn from a zero-centred Normal distribution) might be off. Some reasons it might be reasonable to omit the (1|A) term:




                          • there is something about your data that would eliminate any differences in A (not just experimental control: for example, maybe you've already subtracted the mean from the observations in each A group)

                          • you get a singular fit, i.e. an estimate of zero variance among levels of A (in this case you'll get exactly the same answer if you drop this term)

                          • you think there are too few levels of A to estimate the variance (in this case it would be more conservative to include A in the model as a fixed effect


                          • if you are using data-driven model selection to decide what terms to include, and the model including (1|A) is not selected (e.g. the AIC is higher). (In general you need to be very careful when doing this kind of "sacrificial pseudoreplication" (sensu Hurlbert 1984) that you don't mess up your coverage/type-I error ...)


                          (Note that the levels of B need to be uniquely coded - otherwise you should specify the nested syntax (1|A/B) (or (1|A) + (1|A:B)) in lme4 as well in nlme ...)






                          share|cite|improve this answer












                          In general I would recommend including both (1|A) and (1|B) in the model - otherwise you're not fitting the maximal model, i.e. the model that includes all terms that are statistically identifiable. If the variation in A is reasonably large and you omit the (1|A) term, the model for the random effects corresponding to the remaining (1|B) term (that they are independently drawn from a zero-centred Normal distribution) might be off. Some reasons it might be reasonable to omit the (1|A) term:




                          • there is something about your data that would eliminate any differences in A (not just experimental control: for example, maybe you've already subtracted the mean from the observations in each A group)

                          • you get a singular fit, i.e. an estimate of zero variance among levels of A (in this case you'll get exactly the same answer if you drop this term)

                          • you think there are too few levels of A to estimate the variance (in this case it would be more conservative to include A in the model as a fixed effect


                          • if you are using data-driven model selection to decide what terms to include, and the model including (1|A) is not selected (e.g. the AIC is higher). (In general you need to be very careful when doing this kind of "sacrificial pseudoreplication" (sensu Hurlbert 1984) that you don't mess up your coverage/type-I error ...)


                          (Note that the levels of B need to be uniquely coded - otherwise you should specify the nested syntax (1|A/B) (or (1|A) + (1|A:B)) in lme4 as well in nlme ...)







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 16 at 0:46









                          Ben Bolker

                          22.4k16090




                          22.4k16090






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Cross Validated!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f383156%2fcovariance-structure-with-one-or-more-random-intercepts%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Probability when a professor distributes a quiz and homework assignment to a class of n students.

                              Aardman Animations

                              Are they similar matrix