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
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          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





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          1 Answer
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          1 Answer
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          active

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          active

          oldest

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          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


















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