Probability of observation sequence not knowing previous observations












1












$begingroup$


I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:




Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
enter image description here




I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.



By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.



My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:




    Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
    enter image description here




    I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.



    By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.



    My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:




      Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
      enter image description here




      I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.



      By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.



      My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?










      share|cite|improve this question











      $endgroup$




      I'm trying to study hidden Markov models by solving some exercises, and stumbled upon this one:




      Given HMM 2, what is $P(O_{100} = A, O_{101} = A, O_{102} = A)$?
      enter image description here




      I think that this can be solved using the forward scoring algorithm and by assuming that the previous states can be anything (I'm not sure however if I'm allowed to make this assumption?). This means that the previous steps in the recursion of the forward algorithm are $alpha_A(99) = 1$ and $alpha_B(99) = 1$ with $alpha_i(t)$ where $i$ is one of the states of the HMM and $t$ is the current step in the algorithm.



      By applying $alpha_j(t+1) = (sum_{i=1}^{n}{alpha_i(t) * a_{ij}}) * b_j(O_{t+1})$ up until $O_{102}$ I can find the desired probability.



      My question is now, if I'm making the right assumptions and follow the correct way of solving this, or am I missing something?







      probability probability-theory hidden-markov-models






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 18 at 8:12







      Pieter Verschaffelt

















      asked Jan 18 at 8:06









      Pieter VerschaffeltPieter Verschaffelt

      282110




      282110






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Given: $P(O_1=A)=P(O_1=B)=0.5$



          By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$



          Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$






          share|cite|improve this answer









          $endgroup$














            Your Answer








            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%2f3077958%2fprobability-of-observation-sequence-not-knowing-previous-observations%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Given: $P(O_1=A)=P(O_1=B)=0.5$



            By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$



            Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Given: $P(O_1=A)=P(O_1=B)=0.5$



              By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$



              Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Given: $P(O_1=A)=P(O_1=B)=0.5$



                By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$



                Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$






                share|cite|improve this answer









                $endgroup$



                Given: $P(O_1=A)=P(O_1=B)=0.5$



                By symmetry: $P(O_{100}=A)=P(O_{100}=B)=0.5$



                Thus: $P(O_{100}=A,O_{101}=A,O_{102}=A)=0.5times0.8times0.8=0.32$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 18 at 15:00









                Daniel MathiasDaniel Mathias

                1,41518




                1,41518






























                    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%2f3077958%2fprobability-of-observation-sequence-not-knowing-previous-observations%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

                    Human spaceflight

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    張江高科駅