Joint PDF P[X+Y<=0.5]












1












$begingroup$


I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



I need to find the $P(x+yleq 0.5)$.



For the double integration, I have the following bounds:




  • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


  • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



Here is the initial set up:



$int_0^.5 int_0^{.5-x}(x +y) dydx$.



My steps of integration:



1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






probability







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



    Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



    I need to find the $P(x+yleq 0.5)$.



    For the double integration, I have the following bounds:




    • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


    • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



    Here is the initial set up:



    $int_0^.5 int_0^{.5-x}(x +y) dydx$.



    My steps of integration:



    1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



    2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



    3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



    I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






    probability







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



      Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



      I need to find the $P(x+yleq 0.5)$.



      For the double integration, I have the following bounds:




      • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


      • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



      Here is the initial set up:



      $int_0^.5 int_0^{.5-x}(x +y) dydx$.



      My steps of integration:



      1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



      2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



      3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



      I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






      probability







      share|cite|improve this question











      $endgroup$




      I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:



      Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.



      I need to find the $P(x+yleq 0.5)$.



      For the double integration, I have the following bounds:




      • Outer bound is respectively to $x$ and is from $0$ to $0.5$.


      • Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.



      Here is the initial set up:



      $int_0^.5 int_0^{.5-x}(x +y) dydx$.



      My steps of integration:



      1: $int_0^.5 left(xy + frac{y^2}{2}right)Big|_0^{.5-x}dx$.



      2: $int_0^.5$ $left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx $.



      3: $(frac{x^2}{4} - frac{x^3}{3} + frac{(0.5-x)^3}{6})Big|_0^{0.5}$.



      I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?






      probability




      probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 10:42







      MitterHai

















      asked Dec 5 '18 at 10:19









      MitterHaiMitterHai

      84




      84






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43



















          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40











          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%2f3026902%2fjoint-pdf-pxy-0-5%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









          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43
















          0












          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43














          0












          0








          0





          $begingroup$

          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$






          share|cite|improve this answer









          $endgroup$



          Your made the sign error in step 2 (as indicated by KaviRamaMurthy).



          As an alternative, you can simplify in step 2 before integrating:
          $$int_0^{0.5}left(frac{x}{2} -x^2+frac{(0.5-x)^2}{2}right) dx=
          int_0^{0.5}left(require{cancel}cancel{frac{x}{2}} -x^2+frac{1}{8}-cancel{frac x2}+frac{x^2}{2}right) dx=\
          int_0^{0.5}left(-frac{x^2}{2}+frac18right) dx=
          left(-frac{x^3}{6}+frac x8right)|_0^{0.5}=
          -frac{1}{48}+frac{1}{16}=
          frac 1{24}.$$



          Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $frac1{48}$:
          $$(frac{x^2}{4} - frac{x^3}{3} color{red}{-} frac{(0.5-x)^3}{6})Big|_0^{0.5}=left(frac{1}{16}-frac1{24}-0right)-left(0-0-frac1{48}right)=frac1{24}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 11:22









          farruhotafarruhota

          20k2738




          20k2738












          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43


















          • $begingroup$
            Farruhota, thank you for the awesome feedback. I really appreciate your help.
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:42










          • $begingroup$
            You are welcome. Good luck.
            $endgroup$
            – farruhota
            Dec 5 '18 at 11:43
















          $begingroup$
          Farruhota, thank you for the awesome feedback. I really appreciate your help.
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:42




          $begingroup$
          Farruhota, thank you for the awesome feedback. I really appreciate your help.
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:42












          $begingroup$
          You are welcome. Good luck.
          $endgroup$
          – farruhota
          Dec 5 '18 at 11:43




          $begingroup$
          You are welcome. Good luck.
          $endgroup$
          – farruhota
          Dec 5 '18 at 11:43











          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40
















          1












          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40














          1












          1








          1





          $begingroup$

          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.






          share|cite|improve this answer









          $endgroup$



          Anti-derivative of $frac {(0.5-x)^{2}} 2$ is $-frac {(0.5-x)^{3}} 6$. You are missing the minus sign.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 10:22









          Kavi Rama MurthyKavi Rama Murthy

          56.6k42159




          56.6k42159












          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40


















          • $begingroup$
            Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
            $endgroup$
            – MitterHai
            Dec 5 '18 at 10:38










          • $begingroup$
            @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
            $endgroup$
            – littleO
            Dec 5 '18 at 11:22










          • $begingroup$
            What a face palm moment. Thanks Pro..fe.. I mean littleO!
            $endgroup$
            – MitterHai
            Dec 5 '18 at 11:40
















          $begingroup$
          Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
          $endgroup$
          – MitterHai
          Dec 5 '18 at 10:38




          $begingroup$
          Hi Kavi, thank you for taking on this problem, I figured out how the anti-derivative should have a minus sign, but when you evaluate it at 0.5, doesn't this part of the equation just go to 0?
          $endgroup$
          – MitterHai
          Dec 5 '18 at 10:38












          $begingroup$
          @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
          $endgroup$
          – littleO
          Dec 5 '18 at 11:22




          $begingroup$
          @MitterHai When you evaluate $-(0.5 - x)^3/6$ at $0$, you get a non-zero value. This fixes the problem.
          $endgroup$
          – littleO
          Dec 5 '18 at 11:22












          $begingroup$
          What a face palm moment. Thanks Pro..fe.. I mean littleO!
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:40




          $begingroup$
          What a face palm moment. Thanks Pro..fe.. I mean littleO!
          $endgroup$
          – MitterHai
          Dec 5 '18 at 11:40


















          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%2f3026902%2fjoint-pdf-pxy-0-5%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