How do I apply Law of Large Numbers in this question
$begingroup$
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
asked Jan 8 at 1:18
SABOYSABOY
694311
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 1:18
SABOYSABOY
694311
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 1:18
SABOYSABOY
694311
$endgroup$
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
asked Jan 8 at 1:18
SABOYSABOY
694311
asked Jan 8 at 1:18
SABOYSABOY
694311
asked Jan 8 at 1:18
SABOYSABOY
694311
asked Jan 8 at 1:18
SABOYSABOY
694311
asked Jan 8 at 1:18
SABOYSABOY
694311
694311
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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.
answered Jan 8 at 8:31
SongSong
14.1k1633
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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.
answered Jan 8 at 8:31
SongSong
14.1k1633
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 8 at 8:31
SongSong
14.1k1633
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 8 at 8:31
SongSong
14.1k1633
$endgroup$
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.
answered Jan 8 at 8:31
SongSong
14.1k1633
answered Jan 8 at 8:31
SongSong
14.1k1633
answered Jan 8 at 8:31
SongSong
14.1k1633
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
add a comment |
$begingroup$
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
add a comment |
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