How do I apply Law of Large Numbers in this question












0












$begingroup$


Let $K_{0}=1$, and in every round $i =1,...,n$ a player puts all capital $K_{i-1}$ in. A fair coin is thrown. If it falls on Heads he receives $frac{5}{3}$ of his capital outlay, if it falls on Tails then $frac{1}{3}$ of his capital outlay is returned to him.



$1.$ Show $mathbb E[K_{n}]=1$



$2. K_{n} to 0$ a.s., where $n to infty$



Ideas:



$1.$ I am open to corrections, but I think this is fairly straightforward, perhaps even inductive?



$mathbb E[K_{1}]=frac{1}{2}frac{5}{3}K_{0}+frac{1}{2}frac{1}{3}K_{0}=1$



and it goes on $mathbb E[K_{i}]=frac{1}{2}frac{5}{3}E[K_{i-1}]+frac{1}{2}frac{1}{3}E[K_{i-1}]=mathbb E[X_{i-1}]$



Does this suffice?



$2.$



Because we need to show $K_{n} to 0$ a.s., $n to infty$ I assume we need to use the law of large numbers



The only problem is:



$sum_{i=1}^{n}frac{K_{n}}{n}to mathbb E[X_{1}]$, a.s. but as proven in $1.$ $mathbb E[X_{1}]=1$



Oh yes but of course, it is plain to see that $(K_{n})_{n}$ are not independent random variables, and therefore I need to find some smart way to "create" random variables that are independent yet involve $K_{n}$ in order to apply the law of large numbers.



Any ideas/assistance?






real-analysis probability probability-theory random-variables expected-value







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $K_{0}=1$, and in every round $i =1,...,n$ a player puts all capital $K_{i-1}$ in. A fair coin is thrown. If it falls on Heads he receives $frac{5}{3}$ of his capital outlay, if it falls on Tails then $frac{1}{3}$ of his capital outlay is returned to him.



    $1.$ Show $mathbb E[K_{n}]=1$



    $2. K_{n} to 0$ a.s., where $n to infty$



    Ideas:



    $1.$ I am open to corrections, but I think this is fairly straightforward, perhaps even inductive?



    $mathbb E[K_{1}]=frac{1}{2}frac{5}{3}K_{0}+frac{1}{2}frac{1}{3}K_{0}=1$



    and it goes on $mathbb E[K_{i}]=frac{1}{2}frac{5}{3}E[K_{i-1}]+frac{1}{2}frac{1}{3}E[K_{i-1}]=mathbb E[X_{i-1}]$



    Does this suffice?



    $2.$



    Because we need to show $K_{n} to 0$ a.s., $n to infty$ I assume we need to use the law of large numbers



    The only problem is:



    $sum_{i=1}^{n}frac{K_{n}}{n}to mathbb E[X_{1}]$, a.s. but as proven in $1.$ $mathbb E[X_{1}]=1$



    Oh yes but of course, it is plain to see that $(K_{n})_{n}$ are not independent random variables, and therefore I need to find some smart way to "create" random variables that are independent yet involve $K_{n}$ in order to apply the law of large numbers.



    Any ideas/assistance?






    real-analysis probability probability-theory random-variables expected-value







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $K_{0}=1$, and in every round $i =1,...,n$ a player puts all capital $K_{i-1}$ in. A fair coin is thrown. If it falls on Heads he receives $frac{5}{3}$ of his capital outlay, if it falls on Tails then $frac{1}{3}$ of his capital outlay is returned to him.



      $1.$ Show $mathbb E[K_{n}]=1$



      $2. K_{n} to 0$ a.s., where $n to infty$



      Ideas:



      $1.$ I am open to corrections, but I think this is fairly straightforward, perhaps even inductive?



      $mathbb E[K_{1}]=frac{1}{2}frac{5}{3}K_{0}+frac{1}{2}frac{1}{3}K_{0}=1$



      and it goes on $mathbb E[K_{i}]=frac{1}{2}frac{5}{3}E[K_{i-1}]+frac{1}{2}frac{1}{3}E[K_{i-1}]=mathbb E[X_{i-1}]$



      Does this suffice?



      $2.$



      Because we need to show $K_{n} to 0$ a.s., $n to infty$ I assume we need to use the law of large numbers



      The only problem is:



      $sum_{i=1}^{n}frac{K_{n}}{n}to mathbb E[X_{1}]$, a.s. but as proven in $1.$ $mathbb E[X_{1}]=1$



      Oh yes but of course, it is plain to see that $(K_{n})_{n}$ are not independent random variables, and therefore I need to find some smart way to "create" random variables that are independent yet involve $K_{n}$ in order to apply the law of large numbers.



      Any ideas/assistance?






      real-analysis probability probability-theory random-variables expected-value







      share|cite|improve this question









      $endgroup$




      Let $K_{0}=1$, and in every round $i =1,...,n$ a player puts all capital $K_{i-1}$ in. A fair coin is thrown. If it falls on Heads he receives $frac{5}{3}$ of his capital outlay, if it falls on Tails then $frac{1}{3}$ of his capital outlay is returned to him.



      $1.$ Show $mathbb E[K_{n}]=1$



      $2. K_{n} to 0$ a.s., where $n to infty$



      Ideas:



      $1.$ I am open to corrections, but I think this is fairly straightforward, perhaps even inductive?



      $mathbb E[K_{1}]=frac{1}{2}frac{5}{3}K_{0}+frac{1}{2}frac{1}{3}K_{0}=1$



      and it goes on $mathbb E[K_{i}]=frac{1}{2}frac{5}{3}E[K_{i-1}]+frac{1}{2}frac{1}{3}E[K_{i-1}]=mathbb E[X_{i-1}]$



      Does this suffice?



      $2.$



      Because we need to show $K_{n} to 0$ a.s., $n to infty$ I assume we need to use the law of large numbers



      The only problem is:



      $sum_{i=1}^{n}frac{K_{n}}{n}to mathbb E[X_{1}]$, a.s. but as proven in $1.$ $mathbb E[X_{1}]=1$



      Oh yes but of course, it is plain to see that $(K_{n})_{n}$ are not independent random variables, and therefore I need to find some smart way to "create" random variables that are independent yet involve $K_{n}$ in order to apply the law of large numbers.



      Any ideas/assistance?






      real-analysis probability probability-theory random-variables expected-value




      real-analysis probability probability-theory random-variables expected-value






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 1:18









      SABOYSABOY

      694311




      694311






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          In this game, what is independent is the rate of the return $frac{K_{n}}{K_{n-1}}$. And it is a function of $X_n$, the result of the $n$-th round coin toss, $X_n = 1$ if coin falls in head and $0$ otherwise. We can see that $frac{K_{n}}{K_{n-1}} = left(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}$ and
          $$
          K_n = 1cdotleft(frac{5}{3}right)^{X_1}left(frac{1}{3}right)^{1-X_1}cdotsleft(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}=left(frac{5}{3}right)^{S_n}left(frac{1}{3}right)^{n-S_n}
          $$           where $S_n= sum_{j=1}^n X_j$ is the total number of heads until the $n$-th round. Take log on both sides and we have
          $$
          frac{1}{n}log K_n = frac{S_n}{n}logleft(frac{5}{3}right)+left(1-frac{S_n}{n}right)logleft(frac{1}{3}right).
          $$           Strong law of large numbers implies $frac{S_n}{n}to E[X_1] =frac{1}{2}$ almost surely. Thus, it holds
          $$
          frac{1}{n}log K_n to frac{1}{2}logleft(frac{5}{9}right)=c<0.
          $$           It says that $K_n^{frac{1}{n}}to e^c <1 $ and hence $K_n to 0$ almost surely.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, but why is the rate of return independent, surely it depends on $K_{i}-1$ in the formula $Y_{i}:=frac{K_{i}}{K_{i-1}}$ and similarly $K_{i-1}$ is present in $Y_{i-1}:=frac{K_{i-1}}{K_{i-2}}$
            $endgroup$
            – SABOY
            Jan 8 at 9:38












          • $begingroup$
            Well, yes it's true that $Y_i=frac{K_i}{K_{i-1}}$ contains $K_{i-1}$ in the expression. But it is also true that $Y_i$ and $K_{i-1}$ are independent. Why the previous fact does not contradict the latter independence statement? The reason is that $Y_i$ only depends on the $i$-th round coin toss $X_i$, which is (intuitively) independent of $K_{i-1}$. On the other hand, $K_i$ is a variable dependent on the previous asset $K_{i-1}$. This dependence is eliminated by factoring $K_i$ by $K_{i-1}$. This (which is equivalent to take logarithm and then difference) is one way of transforming the given
            $endgroup$
            – Song
            Jan 8 at 9:51












          • $begingroup$
            data to get new data which are more easy to deal with.
            $endgroup$
            – Song
            Jan 8 at 9:54











          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%2f3065708%2fhow-do-i-apply-law-of-large-numbers-in-this-question%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          In this game, what is independent is the rate of the return $frac{K_{n}}{K_{n-1}}$. And it is a function of $X_n$, the result of the $n$-th round coin toss, $X_n = 1$ if coin falls in head and $0$ otherwise. We can see that $frac{K_{n}}{K_{n-1}} = left(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}$ and
          $$
          K_n = 1cdotleft(frac{5}{3}right)^{X_1}left(frac{1}{3}right)^{1-X_1}cdotsleft(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}=left(frac{5}{3}right)^{S_n}left(frac{1}{3}right)^{n-S_n}
          $$           where $S_n= sum_{j=1}^n X_j$ is the total number of heads until the $n$-th round. Take log on both sides and we have
          $$
          frac{1}{n}log K_n = frac{S_n}{n}logleft(frac{5}{3}right)+left(1-frac{S_n}{n}right)logleft(frac{1}{3}right).
          $$           Strong law of large numbers implies $frac{S_n}{n}to E[X_1] =frac{1}{2}$ almost surely. Thus, it holds
          $$
          frac{1}{n}log K_n to frac{1}{2}logleft(frac{5}{9}right)=c<0.
          $$           It says that $K_n^{frac{1}{n}}to e^c <1 $ and hence $K_n to 0$ almost surely.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, but why is the rate of return independent, surely it depends on $K_{i}-1$ in the formula $Y_{i}:=frac{K_{i}}{K_{i-1}}$ and similarly $K_{i-1}$ is present in $Y_{i-1}:=frac{K_{i-1}}{K_{i-2}}$
            $endgroup$
            – SABOY
            Jan 8 at 9:38












          • $begingroup$
            Well, yes it's true that $Y_i=frac{K_i}{K_{i-1}}$ contains $K_{i-1}$ in the expression. But it is also true that $Y_i$ and $K_{i-1}$ are independent. Why the previous fact does not contradict the latter independence statement? The reason is that $Y_i$ only depends on the $i$-th round coin toss $X_i$, which is (intuitively) independent of $K_{i-1}$. On the other hand, $K_i$ is a variable dependent on the previous asset $K_{i-1}$. This dependence is eliminated by factoring $K_i$ by $K_{i-1}$. This (which is equivalent to take logarithm and then difference) is one way of transforming the given
            $endgroup$
            – Song
            Jan 8 at 9:51












          • $begingroup$
            data to get new data which are more easy to deal with.
            $endgroup$
            – Song
            Jan 8 at 9:54
















          2












          $begingroup$

          In this game, what is independent is the rate of the return $frac{K_{n}}{K_{n-1}}$. And it is a function of $X_n$, the result of the $n$-th round coin toss, $X_n = 1$ if coin falls in head and $0$ otherwise. We can see that $frac{K_{n}}{K_{n-1}} = left(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}$ and
          $$
          K_n = 1cdotleft(frac{5}{3}right)^{X_1}left(frac{1}{3}right)^{1-X_1}cdotsleft(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}=left(frac{5}{3}right)^{S_n}left(frac{1}{3}right)^{n-S_n}
          $$           where $S_n= sum_{j=1}^n X_j$ is the total number of heads until the $n$-th round. Take log on both sides and we have
          $$
          frac{1}{n}log K_n = frac{S_n}{n}logleft(frac{5}{3}right)+left(1-frac{S_n}{n}right)logleft(frac{1}{3}right).
          $$           Strong law of large numbers implies $frac{S_n}{n}to E[X_1] =frac{1}{2}$ almost surely. Thus, it holds
          $$
          frac{1}{n}log K_n to frac{1}{2}logleft(frac{5}{9}right)=c<0.
          $$           It says that $K_n^{frac{1}{n}}to e^c <1 $ and hence $K_n to 0$ almost surely.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you, but why is the rate of return independent, surely it depends on $K_{i}-1$ in the formula $Y_{i}:=frac{K_{i}}{K_{i-1}}$ and similarly $K_{i-1}$ is present in $Y_{i-1}:=frac{K_{i-1}}{K_{i-2}}$
            $endgroup$
            – SABOY
            Jan 8 at 9:38












          • $begingroup$
            Well, yes it's true that $Y_i=frac{K_i}{K_{i-1}}$ contains $K_{i-1}$ in the expression. But it is also true that $Y_i$ and $K_{i-1}$ are independent. Why the previous fact does not contradict the latter independence statement? The reason is that $Y_i$ only depends on the $i$-th round coin toss $X_i$, which is (intuitively) independent of $K_{i-1}$. On the other hand, $K_i$ is a variable dependent on the previous asset $K_{i-1}$. This dependence is eliminated by factoring $K_i$ by $K_{i-1}$. This (which is equivalent to take logarithm and then difference) is one way of transforming the given
            $endgroup$
            – Song
            Jan 8 at 9:51












          • $begingroup$
            data to get new data which are more easy to deal with.
            $endgroup$
            – Song
            Jan 8 at 9:54














          2












          2








          2





          $begingroup$

          In this game, what is independent is the rate of the return $frac{K_{n}}{K_{n-1}}$. And it is a function of $X_n$, the result of the $n$-th round coin toss, $X_n = 1$ if coin falls in head and $0$ otherwise. We can see that $frac{K_{n}}{K_{n-1}} = left(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}$ and
          $$
          K_n = 1cdotleft(frac{5}{3}right)^{X_1}left(frac{1}{3}right)^{1-X_1}cdotsleft(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}=left(frac{5}{3}right)^{S_n}left(frac{1}{3}right)^{n-S_n}
          $$           where $S_n= sum_{j=1}^n X_j$ is the total number of heads until the $n$-th round. Take log on both sides and we have
          $$
          frac{1}{n}log K_n = frac{S_n}{n}logleft(frac{5}{3}right)+left(1-frac{S_n}{n}right)logleft(frac{1}{3}right).
          $$           Strong law of large numbers implies $frac{S_n}{n}to E[X_1] =frac{1}{2}$ almost surely. Thus, it holds
          $$
          frac{1}{n}log K_n to frac{1}{2}logleft(frac{5}{9}right)=c<0.
          $$           It says that $K_n^{frac{1}{n}}to e^c <1 $ and hence $K_n to 0$ almost surely.






          share|cite|improve this answer









          $endgroup$



          In this game, what is independent is the rate of the return $frac{K_{n}}{K_{n-1}}$. And it is a function of $X_n$, the result of the $n$-th round coin toss, $X_n = 1$ if coin falls in head and $0$ otherwise. We can see that $frac{K_{n}}{K_{n-1}} = left(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}$ and
          $$
          K_n = 1cdotleft(frac{5}{3}right)^{X_1}left(frac{1}{3}right)^{1-X_1}cdotsleft(frac{5}{3}right)^{X_n}left(frac{1}{3}right)^{1-X_n}=left(frac{5}{3}right)^{S_n}left(frac{1}{3}right)^{n-S_n}
          $$           where $S_n= sum_{j=1}^n X_j$ is the total number of heads until the $n$-th round. Take log on both sides and we have
          $$
          frac{1}{n}log K_n = frac{S_n}{n}logleft(frac{5}{3}right)+left(1-frac{S_n}{n}right)logleft(frac{1}{3}right).
          $$           Strong law of large numbers implies $frac{S_n}{n}to E[X_1] =frac{1}{2}$ almost surely. Thus, it holds
          $$
          frac{1}{n}log K_n to frac{1}{2}logleft(frac{5}{9}right)=c<0.
          $$           It says that $K_n^{frac{1}{n}}to e^c <1 $ and hence $K_n to 0$ almost surely.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 8:31









          SongSong

          14.1k1633




          14.1k1633












          • $begingroup$
            Thank you, but why is the rate of return independent, surely it depends on $K_{i}-1$ in the formula $Y_{i}:=frac{K_{i}}{K_{i-1}}$ and similarly $K_{i-1}$ is present in $Y_{i-1}:=frac{K_{i-1}}{K_{i-2}}$
            $endgroup$
            – SABOY
            Jan 8 at 9:38












          • $begingroup$
            Well, yes it's true that $Y_i=frac{K_i}{K_{i-1}}$ contains $K_{i-1}$ in the expression. But it is also true that $Y_i$ and $K_{i-1}$ are independent. Why the previous fact does not contradict the latter independence statement? The reason is that $Y_i$ only depends on the $i$-th round coin toss $X_i$, which is (intuitively) independent of $K_{i-1}$. On the other hand, $K_i$ is a variable dependent on the previous asset $K_{i-1}$. This dependence is eliminated by factoring $K_i$ by $K_{i-1}$. This (which is equivalent to take logarithm and then difference) is one way of transforming the given
            $endgroup$
            – Song
            Jan 8 at 9:51












          • $begingroup$
            data to get new data which are more easy to deal with.
            $endgroup$
            – Song
            Jan 8 at 9:54


















          • $begingroup$
            Thank you, but why is the rate of return independent, surely it depends on $K_{i}-1$ in the formula $Y_{i}:=frac{K_{i}}{K_{i-1}}$ and similarly $K_{i-1}$ is present in $Y_{i-1}:=frac{K_{i-1}}{K_{i-2}}$
            $endgroup$
            – SABOY
            Jan 8 at 9:38












          • $begingroup$
            Well, yes it's true that $Y_i=frac{K_i}{K_{i-1}}$ contains $K_{i-1}$ in the expression. But it is also true that $Y_i$ and $K_{i-1}$ are independent. Why the previous fact does not contradict the latter independence statement? The reason is that $Y_i$ only depends on the $i$-th round coin toss $X_i$, which is (intuitively) independent of $K_{i-1}$. On the other hand, $K_i$ is a variable dependent on the previous asset $K_{i-1}$. This dependence is eliminated by factoring $K_i$ by $K_{i-1}$. This (which is equivalent to take logarithm and then difference) is one way of transforming the given
            $endgroup$
            – Song
            Jan 8 at 9:51












          • $begingroup$
            data to get new data which are more easy to deal with.
            $endgroup$
            – Song
            Jan 8 at 9:54
















          $begingroup$
          Thank you, but why is the rate of return independent, surely it depends on $K_{i}-1$ in the formula $Y_{i}:=frac{K_{i}}{K_{i-1}}$ and similarly $K_{i-1}$ is present in $Y_{i-1}:=frac{K_{i-1}}{K_{i-2}}$
          $endgroup$
          – SABOY
          Jan 8 at 9:38






          $begingroup$
          Thank you, but why is the rate of return independent, surely it depends on $K_{i}-1$ in the formula $Y_{i}:=frac{K_{i}}{K_{i-1}}$ and similarly $K_{i-1}$ is present in $Y_{i-1}:=frac{K_{i-1}}{K_{i-2}}$
          $endgroup$
          – SABOY
          Jan 8 at 9:38














          $begingroup$
          Well, yes it's true that $Y_i=frac{K_i}{K_{i-1}}$ contains $K_{i-1}$ in the expression. But it is also true that $Y_i$ and $K_{i-1}$ are independent. Why the previous fact does not contradict the latter independence statement? The reason is that $Y_i$ only depends on the $i$-th round coin toss $X_i$, which is (intuitively) independent of $K_{i-1}$. On the other hand, $K_i$ is a variable dependent on the previous asset $K_{i-1}$. This dependence is eliminated by factoring $K_i$ by $K_{i-1}$. This (which is equivalent to take logarithm and then difference) is one way of transforming the given
          $endgroup$
          – Song
          Jan 8 at 9:51






          $begingroup$
          Well, yes it's true that $Y_i=frac{K_i}{K_{i-1}}$ contains $K_{i-1}$ in the expression. But it is also true that $Y_i$ and $K_{i-1}$ are independent. Why the previous fact does not contradict the latter independence statement? The reason is that $Y_i$ only depends on the $i$-th round coin toss $X_i$, which is (intuitively) independent of $K_{i-1}$. On the other hand, $K_i$ is a variable dependent on the previous asset $K_{i-1}$. This dependence is eliminated by factoring $K_i$ by $K_{i-1}$. This (which is equivalent to take logarithm and then difference) is one way of transforming the given
          $endgroup$
          – Song
          Jan 8 at 9:51














          $begingroup$
          data to get new data which are more easy to deal with.
          $endgroup$
          – Song
          Jan 8 at 9:54




          $begingroup$
          data to get new data which are more easy to deal with.
          $endgroup$
          – Song
          Jan 8 at 9:54


















          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%2f3065708%2fhow-do-i-apply-law-of-large-numbers-in-this-question%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Human spaceflight

          Image
          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
          Read more

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

          Image
          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
          Read more

          File:DeusFollowingSea.jpg

          Image
          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
          Read more