4Derivation:

mt+1∕pt+1 = (m(t − ct)R∕)pt+1 + yt+1∕pt+1
     m    =  mt-−  ct R -pt-+ yt+1
      t+1     pt   pt   pt+1  pt+1
          = (mt − ct)(R∕𝒢t+1)+ 𝜃t+1.
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               at