Why does finding the partial derivative solve this payoff matrix?












0












$begingroup$


So I'm taking this math course and I decided to do all the homework the first day of class, and as I was pulling an all-nighter doing it, I got to the topic of game theory and payoff matrices. I never took a formal course in multi-variable calculus, but I learned on my own about partial derivatives and I get the gist of it. I got to this question on the homework:
Question



Well in the textbook, it calls for setting up the rows choices as [x, 1-x] and having the column's choices being [0,1] and then repeating the process with [1,0] to find the optimum row strategy, and then doing the inverse with columns. When I saw how tedious this would be, I noticed that there're only two unknowns, so I could just have an x and y and use some multi-variable calculus.



I simply did this:
matmul



and did dp/dx and dp/dy and set each equal to 0 and solved for x and y. I plugged these into the problem and it was the right answer.



Now looking back on this, I can't seem to figure out why this actually solves the problem. I think I don't have a very good understanding of partial derivatives, but from standard Calc 1, I know you would set the derivative to 0 and that would give you the min/max of the function. But translating this to 2 variables is confusing to me, and consequently I've spent the past hour trying to understand why this solves the problem in the first place. Could somebody explain to me very simply how this is actually working and why this finds the optimum strategies?



Thanks!!






matrices multivariable-calculus optimization partial-derivative game-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    So I'm taking this math course and I decided to do all the homework the first day of class, and as I was pulling an all-nighter doing it, I got to the topic of game theory and payoff matrices. I never took a formal course in multi-variable calculus, but I learned on my own about partial derivatives and I get the gist of it. I got to this question on the homework:
    Question



    Well in the textbook, it calls for setting up the rows choices as [x, 1-x] and having the column's choices being [0,1] and then repeating the process with [1,0] to find the optimum row strategy, and then doing the inverse with columns. When I saw how tedious this would be, I noticed that there're only two unknowns, so I could just have an x and y and use some multi-variable calculus.



    I simply did this:
    matmul



    and did dp/dx and dp/dy and set each equal to 0 and solved for x and y. I plugged these into the problem and it was the right answer.



    Now looking back on this, I can't seem to figure out why this actually solves the problem. I think I don't have a very good understanding of partial derivatives, but from standard Calc 1, I know you would set the derivative to 0 and that would give you the min/max of the function. But translating this to 2 variables is confusing to me, and consequently I've spent the past hour trying to understand why this solves the problem in the first place. Could somebody explain to me very simply how this is actually working and why this finds the optimum strategies?



    Thanks!!






    matrices multivariable-calculus optimization partial-derivative game-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      So I'm taking this math course and I decided to do all the homework the first day of class, and as I was pulling an all-nighter doing it, I got to the topic of game theory and payoff matrices. I never took a formal course in multi-variable calculus, but I learned on my own about partial derivatives and I get the gist of it. I got to this question on the homework:
      Question



      Well in the textbook, it calls for setting up the rows choices as [x, 1-x] and having the column's choices being [0,1] and then repeating the process with [1,0] to find the optimum row strategy, and then doing the inverse with columns. When I saw how tedious this would be, I noticed that there're only two unknowns, so I could just have an x and y and use some multi-variable calculus.



      I simply did this:
      matmul



      and did dp/dx and dp/dy and set each equal to 0 and solved for x and y. I plugged these into the problem and it was the right answer.



      Now looking back on this, I can't seem to figure out why this actually solves the problem. I think I don't have a very good understanding of partial derivatives, but from standard Calc 1, I know you would set the derivative to 0 and that would give you the min/max of the function. But translating this to 2 variables is confusing to me, and consequently I've spent the past hour trying to understand why this solves the problem in the first place. Could somebody explain to me very simply how this is actually working and why this finds the optimum strategies?



      Thanks!!






      matrices multivariable-calculus optimization partial-derivative game-theory







      share|cite|improve this question









      $endgroup$




      So I'm taking this math course and I decided to do all the homework the first day of class, and as I was pulling an all-nighter doing it, I got to the topic of game theory and payoff matrices. I never took a formal course in multi-variable calculus, but I learned on my own about partial derivatives and I get the gist of it. I got to this question on the homework:
      Question



      Well in the textbook, it calls for setting up the rows choices as [x, 1-x] and having the column's choices being [0,1] and then repeating the process with [1,0] to find the optimum row strategy, and then doing the inverse with columns. When I saw how tedious this would be, I noticed that there're only two unknowns, so I could just have an x and y and use some multi-variable calculus.



      I simply did this:
      matmul



      and did dp/dx and dp/dy and set each equal to 0 and solved for x and y. I plugged these into the problem and it was the right answer.



      Now looking back on this, I can't seem to figure out why this actually solves the problem. I think I don't have a very good understanding of partial derivatives, but from standard Calc 1, I know you would set the derivative to 0 and that would give you the min/max of the function. But translating this to 2 variables is confusing to me, and consequently I've spent the past hour trying to understand why this solves the problem in the first place. Could somebody explain to me very simply how this is actually working and why this finds the optimum strategies?



      Thanks!!






      matrices multivariable-calculus optimization partial-derivative game-theory




      matrices multivariable-calculus optimization partial-derivative game-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 17 at 5:05









      Carson PCarson P

      11




      11






















          0






          active

          oldest

          votes











          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: "69"
          };
          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: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          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
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076612%2fwhy-does-finding-the-partial-derivative-solve-this-payoff-matrix%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          0






          active

          oldest

          votes








          0






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • 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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076612%2fwhy-does-finding-the-partial-derivative-solve-this-payoff-matrix%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

          Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

          Image
          28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
          Read more

          Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

          Image
          up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
          Read more

          A Topological Invariant for $pi_3(U(n))$

          Image
          3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
          Read more