Solving $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}$












0












$begingroup$


I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23
















0












$begingroup$


I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23














0












0








0





$begingroup$


I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving







share|cite|improve this question









$endgroup$




I am trying to solve the recurrence relation



$$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.$$



The context of this recurrence relation is as follows: if I start with $20, and I win $1 for every head and lose $1 for every tail, I obtain the above recurrence relation for $p$. I lose the overall game if I lose all my money, and win the overall game if I accumulate $100.



Now, I understand that substituting $p_k = p^k$ is the usual approach, which gives $p=1$. While clearly this is a solution, in the context of this above game something’s not right – where in my logic have I gone wrong? Thank you!






probability recurrence-relations problem-solving




probability recurrence-relations problem-solving






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 3:51









user107224user107224

463314




463314








  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23














  • 1




    $begingroup$
    What do you mean by p?
    $endgroup$
    – Martund
    Dec 20 '18 at 3:55












  • $begingroup$
    This may help.
    $endgroup$
    – Jam
    Dec 20 '18 at 4:04










  • $begingroup$
    The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
    $endgroup$
    – Michael Burr
    Dec 20 '18 at 4:23








1




1




$begingroup$
What do you mean by p?
$endgroup$
– Martund
Dec 20 '18 at 3:55






$begingroup$
What do you mean by p?
$endgroup$
– Martund
Dec 20 '18 at 3:55














$begingroup$
This may help.
$endgroup$
– Jam
Dec 20 '18 at 4:04




$begingroup$
This may help.
$endgroup$
– Jam
Dec 20 '18 at 4:04












$begingroup$
The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
$endgroup$
– Michael Burr
Dec 20 '18 at 4:23




$begingroup$
The relationship between your recurrence relation and the stated problem is not clear. What do these $p_k$'s represent?
$endgroup$
– Michael Burr
Dec 20 '18 at 4:23










3 Answers
3






active

oldest

votes


















2












$begingroup$

Write
$p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
$
in the form
$p_{k+1}
=2p_k-p_{k-1}
$.



This has characteristic polynomial
$d^2 = 2d-1$
or
$d^2-2d+1 = 0$
or
$(d-1)^2 = 0$.



The repeated root means that
the two solutions are
$r^k$
and $kr^k$.



The $r^k$ solution gives
$r^{k+1}
=2r^k-r^{k-1}
$
or
$r^2
=2r-1
$,
so $r = 1$
as you got.



The $kr^k$ solution gives,
since $r=1$,
just $k$.



To check, if
$p_k = k$
then
$2p_k-p_{k-1}
=2k-(k-1)
=k+1
=p_{k+1}
$.



Therefore the solutions are
$1$ and $k$,
so the general solution is
$a + bk$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
    $endgroup$
    – Ankit Kumar
    Dec 20 '18 at 5:00












  • $begingroup$
    I forgot about the repeated root, thank you!
    $endgroup$
    – user107224
    Dec 20 '18 at 9:47



















0












$begingroup$

If $p_0$ and $p_1$ are given, then an easy proof by induction gives



$$p_n=n(p_1-p_0)+p_0.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    A simple argument lets us avoid solving a recurrence relation.



    The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



    (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






    share|cite|improve this answer









    $endgroup$













      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%2f3047133%2fsolving-p-k-frac12p-k1-frac12p-k-1%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47
















      2












      $begingroup$

      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47














      2












      2








      2





      $begingroup$

      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.






      share|cite|improve this answer









      $endgroup$



      Write
      $p_k=frac{1}{2}p_{k+1} + frac{1}{2}p_{k-1}.
      $
      in the form
      $p_{k+1}
      =2p_k-p_{k-1}
      $      .



      This has characteristic polynomial
      $d^2 = 2d-1$
      or
      $d^2-2d+1 = 0$
      or
      $(d-1)^2 = 0$.



      The repeated root means that
      the two solutions are
      $r^k$
      and $kr^k$.



      The $r^k$ solution gives
      $r^{k+1}
      =2r^k-r^{k-1}
      $
      or
      $r^2
      =2r-1
      $      ,
      so $r = 1$
      as you got.



      The $kr^k$ solution gives,
      since $r=1$,
      just $k$.



      To check, if
      $p_k = k$
      then
      $2p_k-p_{k-1}
      =2k-(k-1)
      =k+1
      =p_{k+1}
      $      .



      Therefore the solutions are
      $1$ and $k$,
      so the general solution is
      $a + bk$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 20 '18 at 4:17









      marty cohenmarty cohen

      74.5k549129




      74.5k549129








      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47














      • 1




        $begingroup$
        Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
        $endgroup$
        – Ankit Kumar
        Dec 20 '18 at 5:00












      • $begingroup$
        I forgot about the repeated root, thank you!
        $endgroup$
        – user107224
        Dec 20 '18 at 9:47








      1




      1




      $begingroup$
      Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
      $endgroup$
      – Ankit Kumar
      Dec 20 '18 at 5:00






      $begingroup$
      Perfect answer. I'd like to add just one thing. $(d-a)^2$ itself gives $a^k$ and $ka^k$ ($a=1$ in this case). You need not put r and evaluate it separately.
      $endgroup$
      – Ankit Kumar
      Dec 20 '18 at 5:00














      $begingroup$
      I forgot about the repeated root, thank you!
      $endgroup$
      – user107224
      Dec 20 '18 at 9:47




      $begingroup$
      I forgot about the repeated root, thank you!
      $endgroup$
      – user107224
      Dec 20 '18 at 9:47











      0












      $begingroup$

      If $p_0$ and $p_1$ are given, then an easy proof by induction gives



      $$p_n=n(p_1-p_0)+p_0.$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        If $p_0$ and $p_1$ are given, then an easy proof by induction gives



        $$p_n=n(p_1-p_0)+p_0.$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          If $p_0$ and $p_1$ are given, then an easy proof by induction gives



          $$p_n=n(p_1-p_0)+p_0.$$






          share|cite|improve this answer









          $endgroup$



          If $p_0$ and $p_1$ are given, then an easy proof by induction gives



          $$p_n=n(p_1-p_0)+p_0.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 4:19









          FredFred

          48.6k11849




          48.6k11849























              0












              $begingroup$

              A simple argument lets us avoid solving a recurrence relation.



              The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



              (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                A simple argument lets us avoid solving a recurrence relation.



                The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



                (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  A simple argument lets us avoid solving a recurrence relation.



                  The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



                  (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)






                  share|cite|improve this answer









                  $endgroup$



                  A simple argument lets us avoid solving a recurrence relation.



                  The game is fair, so the expected amount of money you end with is equal to the money you start with. You begin with $20; you end with $100 with some unknown probability $x$, and $0 otherwise. Therefore $20 = 100x$, which yields $x = frac15$.



                  (Formally, we should check the hypotheses of Wald's identity to know that the expected amount of money you end with is equal to the money you start with. This means we want the number of steps in the game to be a stopping time (which it is: whether you have won or lost after $n$ games depends only on the outcomes of those $n$ games) and finite with probability $1$ (which is true in any finite Markov chain), while the bets should be bounded in addition to being fair.)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 4:45









                  Misha LavrovMisha Lavrov

                  47.8k657107




                  47.8k657107






























                      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%2f3047133%2fsolving-p-k-frac12p-k1-frac12p-k-1%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

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more