Random Variable Transformation normal/binomial












0












$begingroup$


I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10
















0












$begingroup$


I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10














0












0








0





$begingroup$


I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables







share|cite|improve this question











$endgroup$




I have the following problem I can not solve:



We have two indipendent random Variables given by:
$$
X sim N_{(0,1)}
$$
and
$$
Y_p sim B_{(1,p)}
$$
Now I want to show, that $Z_p sim N_{(0,1)}$ $forall p in (0,1)$ , with
$$
Z_p = (-1)^{Y_p}cdot X
$$
Additionally there is a Hint whoch says, that:
$$
mathbb{P}(A) = mathbb{P}(A cap { Y_p = 1}) + mathbb{P}(A cap { Y_p = 0})
$$



I know that a transformation is given by:
$$
F_Z(z) = mathbb{P}(Z_p < z) = mathbb{P}((-1)^{Y_p}cdot X < z)
$$
And this I somehow get my boundarys for the integration. But I cannot figure out how this is done.






random-variables




random-variables






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 17:44







RedCrayon

















asked Dec 16 '18 at 15:47









RedCrayonRedCrayon

62




62












  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10


















  • $begingroup$
    You should add to your question that $X$ and $Y_p$ are independent.
    $endgroup$
    – drhab
    Dec 16 '18 at 16:10
















$begingroup$
You should add to your question that $X$ and $Y_p$ are independent.
$endgroup$
– drhab
Dec 16 '18 at 16:10




$begingroup$
You should add to your question that $X$ and $Y_p$ are independent.
$endgroup$
– drhab
Dec 16 '18 at 16:10










2 Answers
2






active

oldest

votes


















1












$begingroup$

I preassume that $X$ and $Y_p$ are independent.



For every Borel measurable $A$:
$$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
& =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
& =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
& =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
& =Pleft(Xin Aright)
end{aligned}
$$



The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Welcome to MSE! :-)



    First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



    Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
    begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
    The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
    begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
    The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
    $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
    Hence, $Z_p sim mathrm{N}(0,1)$.






    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%2f3042748%2frandom-variable-transformation-normal-binomial%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









      1












      $begingroup$

      I preassume that $X$ and $Y_p$ are independent.



      For every Borel measurable $A$:
      $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
      & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
      & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
      & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
      & =Pleft(Xin Aright)
      end{aligned}
      $$



      The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        I preassume that $X$ and $Y_p$ are independent.



        For every Borel measurable $A$:
        $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
        & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
        & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
        & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
        & =Pleft(Xin Aright)
        end{aligned}
        $$



        The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          I preassume that $X$ and $Y_p$ are independent.



          For every Borel measurable $A$:
          $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
          & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
          & =Pleft(Xin Aright)
          end{aligned}
          $$



          The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.






          share|cite|improve this answer









          $endgroup$



          I preassume that $X$ and $Y_p$ are independent.



          For every Borel measurable $A$:
          $$begin{aligned}Pleft(Z_{p}in Aright) & =Pleft(Z_{p}in Amid Y_{p}=0right)Pleft(Y_{p}=0right)+Pleft(Z_{p}in Amid Y_{p}=1right)Pleft(Y_{p}=1right)\
          & =Pleft(Xin Amid Y_{p}=0right)left(1-pright)+Pleft(-Xin Amid Y_{p}=1right)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(-Xin Aright)p\
          & =Pleft(Xin Aright)left(1-pright)+Pleft(Xin Aright)p\
          & =Pleft(Xin Aright)
          end{aligned}
          $$



          The third equality rests on independence and the fourth equality rests on the fact that $X$ and $-X$ have the same distribution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 16:08









          drhabdrhab

          103k545136




          103k545136























              1












              $begingroup$

              Welcome to MSE! :-)



              First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



              Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
              begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
              The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
              begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
              The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
              $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
              Hence, $Z_p sim mathrm{N}(0,1)$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Welcome to MSE! :-)



                First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



                Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
                begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
                The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
                begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
                The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
                $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
                Hence, $Z_p sim mathrm{N}(0,1)$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Welcome to MSE! :-)



                  First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



                  Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
                  begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
                  The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
                  begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
                  The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
                  $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
                  Hence, $Z_p sim mathrm{N}(0,1)$.






                  share|cite|improve this answer









                  $endgroup$



                  Welcome to MSE! :-)



                  First note, that if $X sim mathrm{N}(0,1)$, that also $-X sim mathrm{N}(0,1)$, you can see this when noting, that the probability density function of $mathrm{N}(0,1)$ is symmetric around $0$. Which to me would be an intuitive explanation of why this statement is true. However, this explanation works only, if $X$ and $Y_p$ are independent, hence I will assume that in the following.



                  Anyway, let's dig more into detail. Let $A subseteq mathbb{R}$ (measurable) and
                  begin{align*}mathbb{P}(Z_p in A) &= mathbb{P}({Z_p in A} cap {Y_p = 1}) + mathbb{P}({Z_p in A} cap {Y_p = 0}) \ &= mathbb{P}({-X in A} cap {Y_p = 1}) + mathbb{P}({X in A} cap {Y_p = 0}).end{align*}
                  The last statement is true, since we know whether $Z_p = -X$ or $Z_p = X$, if we fix $Y_p$. Since $X$ and $Y_p$ are independent, we obtain
                  begin{align*}mathbb{P}({-X in A} cap {Y_p = 1}) &= pmathbb{P}({-X in A}, \ mathbb{P}({X in A} cap {Y_p = 0}) &= (1-p)mathbb{P}({X in A}.end{align*}
                  The symmetry of $mathrm{N}(0,1)$ implies that $mathbb{P}(X in A) = mathbb{P}(-X in A)$. Applying that first displayed formula, we obtain
                  $$mathbb{P}(Z_p in A) = (p + 1 -p)mathbb{P}(X in A) = mathbb{P}(X in A) = mathrm{N}(0,1)(A).$$
                  Hence, $Z_p sim mathrm{N}(0,1)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 16:08









                  JonasJonas

                  398212




                  398212






























                      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%2f3042748%2frandom-variable-transformation-normal-binomial%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