Let $X_n$ be succession of independent exponentials of parameter $lambda$.












0












$begingroup$


Let $X_n$ be succession of independent exponentials of parameter $lambda$.



Let $Z_n = max(X_1,...,X_n)$



I want to study the convergence of $Z_n$ and of $frac{1}{log{n}}Z_n$



First of all I find the CDF of $Z_n$:
$$ F_{Z_n}(z) = (1-e^{-lambda z})^n $$



The limit of n going to infinity is 0 but there doesn't exist a random function with CDF constant equal to 0, so It doesn't converge in law.



Now if I multiply for $frac{1}{log{n}}$, I get:



$$F_{frac{1}{log{n}}Z_n}(z) = (1- e^{-lambda log{n} z})^n$$



This should converge to the constant $frac{1}{lambda}$ but I am not sure why.






probability weak-convergence







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $X_n$ be succession of independent exponentials of parameter $lambda$.



    Let $Z_n = max(X_1,...,X_n)$



    I want to study the convergence of $Z_n$ and of $frac{1}{log{n}}Z_n$



    First of all I find the CDF of $Z_n$:
    $$ F_{Z_n}(z) = (1-e^{-lambda z})^n $$



    The limit of n going to infinity is 0 but there doesn't exist a random function with CDF constant equal to 0, so It doesn't converge in law.



    Now if I multiply for $frac{1}{log{n}}$, I get:



    $$F_{frac{1}{log{n}}Z_n}(z) = (1- e^{-lambda log{n} z})^n$$



    This should converge to the constant $frac{1}{lambda}$ but I am not sure why.






    probability weak-convergence







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $X_n$ be succession of independent exponentials of parameter $lambda$.



      Let $Z_n = max(X_1,...,X_n)$



      I want to study the convergence of $Z_n$ and of $frac{1}{log{n}}Z_n$



      First of all I find the CDF of $Z_n$:
      $$ F_{Z_n}(z) = (1-e^{-lambda z})^n $$



      The limit of n going to infinity is 0 but there doesn't exist a random function with CDF constant equal to 0, so It doesn't converge in law.



      Now if I multiply for $frac{1}{log{n}}$, I get:



      $$F_{frac{1}{log{n}}Z_n}(z) = (1- e^{-lambda log{n} z})^n$$



      This should converge to the constant $frac{1}{lambda}$ but I am not sure why.






      probability weak-convergence







      share|cite|improve this question









      $endgroup$




      Let $X_n$ be succession of independent exponentials of parameter $lambda$.



      Let $Z_n = max(X_1,...,X_n)$



      I want to study the convergence of $Z_n$ and of $frac{1}{log{n}}Z_n$



      First of all I find the CDF of $Z_n$:
      $$ F_{Z_n}(z) = (1-e^{-lambda z})^n $$



      The limit of n going to infinity is 0 but there doesn't exist a random function with CDF constant equal to 0, so It doesn't converge in law.



      Now if I multiply for $frac{1}{log{n}}$, I get:



      $$F_{frac{1}{log{n}}Z_n}(z) = (1- e^{-lambda log{n} z})^n$$



      This should converge to the constant $frac{1}{lambda}$ but I am not sure why.






      probability weak-convergence




      probability weak-convergence






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 24 at 7:54









      qcc101qcc101

      627213




      627213






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Note that
          begin{align}
          P(Z_nle zlog n) = left(1-frac{1}{n^{lambda z}}right)^n = left(left(1-frac{1}{n^{lambda z}}right)^{n^{lambda z}} right)^{frac{n}{n^{lambda z}}} \
          end{align}
          Therefore,
          begin{align}
          lim_n P(Z_nle zlog n)
          &= frac{1}{e}, quad mbox{ if } lambda z=1\
          &= 1, quad mbox{ if } lambda z>1\
          &= 0, quad mbox{ if } lambda z<1
          end{align}
          which means that the sequence $Z_n/log n$ converges in distribution to $frac{1}{lambda}$ (we do not care about the point $z=1/lambda$ because it is not a point of continuity of the distribution function of the constant random variable $1/lambda$). Since it converges to a constant, $Z_n/log n$ also converges in probability.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I don't see why it should converge to $frac{1}{lambda}$..
            $endgroup$
            – qcc101
            Jan 24 at 9:58










          • $begingroup$
            Think about the CDF of the random variable $1/lambda$ (ie, 0 if $z<1/lambda$ and 1 if $zge 1/lambda$). Then, apply exactly the definition of convergence in distribution (eg, en.wikipedia.org/wiki/… )
            $endgroup$
            – user52227
            Jan 24 at 14:37











          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%2f3085586%2flet-x-n-be-succession-of-independent-exponentials-of-parameter-lambda%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









          1












          $begingroup$

          Note that
          begin{align}
          P(Z_nle zlog n) = left(1-frac{1}{n^{lambda z}}right)^n = left(left(1-frac{1}{n^{lambda z}}right)^{n^{lambda z}} right)^{frac{n}{n^{lambda z}}} \
          end{align}
          Therefore,
          begin{align}
          lim_n P(Z_nle zlog n)
          &= frac{1}{e}, quad mbox{ if } lambda z=1\
          &= 1, quad mbox{ if } lambda z>1\
          &= 0, quad mbox{ if } lambda z<1
          end{align}
          which means that the sequence $Z_n/log n$ converges in distribution to $frac{1}{lambda}$ (we do not care about the point $z=1/lambda$ because it is not a point of continuity of the distribution function of the constant random variable $1/lambda$). Since it converges to a constant, $Z_n/log n$ also converges in probability.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I don't see why it should converge to $frac{1}{lambda}$..
            $endgroup$
            – qcc101
            Jan 24 at 9:58










          • $begingroup$
            Think about the CDF of the random variable $1/lambda$ (ie, 0 if $z<1/lambda$ and 1 if $zge 1/lambda$). Then, apply exactly the definition of convergence in distribution (eg, en.wikipedia.org/wiki/… )
            $endgroup$
            – user52227
            Jan 24 at 14:37
















          1












          $begingroup$

          Note that
          begin{align}
          P(Z_nle zlog n) = left(1-frac{1}{n^{lambda z}}right)^n = left(left(1-frac{1}{n^{lambda z}}right)^{n^{lambda z}} right)^{frac{n}{n^{lambda z}}} \
          end{align}
          Therefore,
          begin{align}
          lim_n P(Z_nle zlog n)
          &= frac{1}{e}, quad mbox{ if } lambda z=1\
          &= 1, quad mbox{ if } lambda z>1\
          &= 0, quad mbox{ if } lambda z<1
          end{align}
          which means that the sequence $Z_n/log n$ converges in distribution to $frac{1}{lambda}$ (we do not care about the point $z=1/lambda$ because it is not a point of continuity of the distribution function of the constant random variable $1/lambda$). Since it converges to a constant, $Z_n/log n$ also converges in probability.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I don't see why it should converge to $frac{1}{lambda}$..
            $endgroup$
            – qcc101
            Jan 24 at 9:58










          • $begingroup$
            Think about the CDF of the random variable $1/lambda$ (ie, 0 if $z<1/lambda$ and 1 if $zge 1/lambda$). Then, apply exactly the definition of convergence in distribution (eg, en.wikipedia.org/wiki/… )
            $endgroup$
            – user52227
            Jan 24 at 14:37














          1












          1








          1





          $begingroup$

          Note that
          begin{align}
          P(Z_nle zlog n) = left(1-frac{1}{n^{lambda z}}right)^n = left(left(1-frac{1}{n^{lambda z}}right)^{n^{lambda z}} right)^{frac{n}{n^{lambda z}}} \
          end{align}
          Therefore,
          begin{align}
          lim_n P(Z_nle zlog n)
          &= frac{1}{e}, quad mbox{ if } lambda z=1\
          &= 1, quad mbox{ if } lambda z>1\
          &= 0, quad mbox{ if } lambda z<1
          end{align}
          which means that the sequence $Z_n/log n$ converges in distribution to $frac{1}{lambda}$ (we do not care about the point $z=1/lambda$ because it is not a point of continuity of the distribution function of the constant random variable $1/lambda$). Since it converges to a constant, $Z_n/log n$ also converges in probability.






          share|cite|improve this answer









          $endgroup$



          Note that
          begin{align}
          P(Z_nle zlog n) = left(1-frac{1}{n^{lambda z}}right)^n = left(left(1-frac{1}{n^{lambda z}}right)^{n^{lambda z}} right)^{frac{n}{n^{lambda z}}} \
          end{align}
          Therefore,
          begin{align}
          lim_n P(Z_nle zlog n)
          &= frac{1}{e}, quad mbox{ if } lambda z=1\
          &= 1, quad mbox{ if } lambda z>1\
          &= 0, quad mbox{ if } lambda z<1
          end{align}
          which means that the sequence $Z_n/log n$ converges in distribution to $frac{1}{lambda}$ (we do not care about the point $z=1/lambda$ because it is not a point of continuity of the distribution function of the constant random variable $1/lambda$). Since it converges to a constant, $Z_n/log n$ also converges in probability.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 8:28









          user52227user52227

          1,038512




          1,038512












          • $begingroup$
            I don't see why it should converge to $frac{1}{lambda}$..
            $endgroup$
            – qcc101
            Jan 24 at 9:58










          • $begingroup$
            Think about the CDF of the random variable $1/lambda$ (ie, 0 if $z<1/lambda$ and 1 if $zge 1/lambda$). Then, apply exactly the definition of convergence in distribution (eg, en.wikipedia.org/wiki/… )
            $endgroup$
            – user52227
            Jan 24 at 14:37


















          • $begingroup$
            I don't see why it should converge to $frac{1}{lambda}$..
            $endgroup$
            – qcc101
            Jan 24 at 9:58










          • $begingroup$
            Think about the CDF of the random variable $1/lambda$ (ie, 0 if $z<1/lambda$ and 1 if $zge 1/lambda$). Then, apply exactly the definition of convergence in distribution (eg, en.wikipedia.org/wiki/… )
            $endgroup$
            – user52227
            Jan 24 at 14:37
















          $begingroup$
          I don't see why it should converge to $frac{1}{lambda}$..
          $endgroup$
          – qcc101
          Jan 24 at 9:58




          $begingroup$
          I don't see why it should converge to $frac{1}{lambda}$..
          $endgroup$
          – qcc101
          Jan 24 at 9:58












          $begingroup$
          Think about the CDF of the random variable $1/lambda$ (ie, 0 if $z<1/lambda$ and 1 if $zge 1/lambda$). Then, apply exactly the definition of convergence in distribution (eg, en.wikipedia.org/wiki/… )
          $endgroup$
          – user52227
          Jan 24 at 14:37




          $begingroup$
          Think about the CDF of the random variable $1/lambda$ (ie, 0 if $z<1/lambda$ and 1 if $zge 1/lambda$). Then, apply exactly the definition of convergence in distribution (eg, en.wikipedia.org/wiki/… )
          $endgroup$
          – user52227
          Jan 24 at 14:37


















          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%2f3085586%2flet-x-n-be-succession-of-independent-exponentials-of-parameter-lambda%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

          MongoDB - Not Authorized To Execute Command

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
          Read more

          How to fix TextFormField cause rebuild widget in Flutter

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
          Read more

          in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

          Image
          1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
          Read more