Expected number of tosses until 3 heads in a row - via Martingale method
$begingroup$
(Quant job interviews - questions and answers - Question 3.8)
For a fair coin, what is the expected number of tosses to get 3 heads in a row
The answer is stated as :
We gamble in such a way that we make money on heads but such that if we get a T on toss $n$, our position is $-n$.
We therefore gamble one unit on the first toss and on each toss after a $T$.
After one head, we gamble three. This guarantees that if we get a $T$ next then we go to $-n$.
After two heads we are therefore up 4, and so we gamble 7 to get us to $-n$ again. our gambling winnings is a martingale since we are making finite trades in a martingale (any bounded trading strategy in a martingale is a martingale).
After three heads our position is $11 - (n-3)=14-n$. the time taken to get three heads is a stopping time with finite expectation so if we stop at it we still have a martingale (Optional sampling theorem) thus $mathbb E (14 -n) = 0 $ and we are done.
I realise there are a few answers to this already, however i am not sure the explanation above, which uses martingale theory, is entirely clear.
The author states that any bounded trading strategy in a martingale is a martingale but what is the underlying martingale here ?
Also I don't understand the underlying motivation for gambling the way described, please could someone put it in more mathematical terms so i can understand the reason for the sizes of the bets ?
probability martingales popular-math
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|
show 1 more comment
$begingroup$
(Quant job interviews - questions and answers - Question 3.8)
For a fair coin, what is the expected number of tosses to get 3 heads in a row
The answer is stated as :
We gamble in such a way that we make money on heads but such that if we get a T on toss $n$, our position is $-n$.
We therefore gamble one unit on the first toss and on each toss after a $T$.
After one head, we gamble three. This guarantees that if we get a $T$ next then we go to $-n$.
After two heads we are therefore up 4, and so we gamble 7 to get us to $-n$ again. our gambling winnings is a martingale since we are making finite trades in a martingale (any bounded trading strategy in a martingale is a martingale).
After three heads our position is $11 - (n-3)=14-n$. the time taken to get three heads is a stopping time with finite expectation so if we stop at it we still have a martingale (Optional sampling theorem) thus $mathbb E (14 -n) = 0 $ and we are done.
I realise there are a few answers to this already, however i am not sure the explanation above, which uses martingale theory, is entirely clear.
The author states that any bounded trading strategy in a martingale is a martingale but what is the underlying martingale here ?
Also I don't understand the underlying motivation for gambling the way described, please could someone put it in more mathematical terms so i can understand the reason for the sizes of the bets ?
probability martingales popular-math
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$begingroup$
Is it clear to you that the betting strategy achieves what the author claims? That is, that if our gambler throws a T on the $n^{th}$ turn (prior to a win) that the net position is $-n$?
$endgroup$
– lulu
Nov 13 '15 at 1:28
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Sorry no it is not :(
$endgroup$
– user3203476
Nov 13 '15 at 1:29
$begingroup$
Ok. Write out several cases (avoiding the string $HHH$). Confirm that it works. Then proceed by induction. The strategy really is the whole point. The bets are specifically structured to ensure that payout.
$endgroup$
– lulu
Nov 13 '15 at 1:30
$begingroup$
The logic of the proof is: "we've given the rules for a betting strategy. From general theory, it is not possible that these rules (nor any other set of rules) has anything other that a $0$ expectation. The beauty of this strategy is that we can actually computed the payout after the expected number of turns.
$endgroup$
– lulu
Nov 13 '15 at 1:33
$begingroup$
This is a really neat way to solve this problem. Consider a smaller scenario - expected time to get one $H$. Bet on $H$ so that if you get $T$ on your $nth$ throw, your position is $-n$. Stop if you get an $H$. In other words, always bet one. If you get an $H$ on the $nth$ throw, your position will be $1-(n-1)=2-n$ and hence $E(n)=2$.
$endgroup$
– A.S.
Nov 13 '15 at 3:52
|
show 1 more comment
$begingroup$
(Quant job interviews - questions and answers - Question 3.8)
For a fair coin, what is the expected number of tosses to get 3 heads in a row
The answer is stated as :
We gamble in such a way that we make money on heads but such that if we get a T on toss $n$, our position is $-n$.
We therefore gamble one unit on the first toss and on each toss after a $T$.
After one head, we gamble three. This guarantees that if we get a $T$ next then we go to $-n$.
After two heads we are therefore up 4, and so we gamble 7 to get us to $-n$ again. our gambling winnings is a martingale since we are making finite trades in a martingale (any bounded trading strategy in a martingale is a martingale).
After three heads our position is $11 - (n-3)=14-n$. the time taken to get three heads is a stopping time with finite expectation so if we stop at it we still have a martingale (Optional sampling theorem) thus $mathbb E (14 -n) = 0 $ and we are done.
I realise there are a few answers to this already, however i am not sure the explanation above, which uses martingale theory, is entirely clear.
The author states that any bounded trading strategy in a martingale is a martingale but what is the underlying martingale here ?
Also I don't understand the underlying motivation for gambling the way described, please could someone put it in more mathematical terms so i can understand the reason for the sizes of the bets ?
probability martingales popular-math
$endgroup$
(Quant job interviews - questions and answers - Question 3.8)
For a fair coin, what is the expected number of tosses to get 3 heads in a row
The answer is stated as :
We gamble in such a way that we make money on heads but such that if we get a T on toss $n$, our position is $-n$.
We therefore gamble one unit on the first toss and on each toss after a $T$.
After one head, we gamble three. This guarantees that if we get a $T$ next then we go to $-n$.
After two heads we are therefore up 4, and so we gamble 7 to get us to $-n$ again. our gambling winnings is a martingale since we are making finite trades in a martingale (any bounded trading strategy in a martingale is a martingale).
After three heads our position is $11 - (n-3)=14-n$. the time taken to get three heads is a stopping time with finite expectation so if we stop at it we still have a martingale (Optional sampling theorem) thus $mathbb E (14 -n) = 0 $ and we are done.
I realise there are a few answers to this already, however i am not sure the explanation above, which uses martingale theory, is entirely clear.
The author states that any bounded trading strategy in a martingale is a martingale but what is the underlying martingale here ?
Also I don't understand the underlying motivation for gambling the way described, please could someone put it in more mathematical terms so i can understand the reason for the sizes of the bets ?
probability martingales popular-math
probability martingales popular-math
edited Nov 13 '15 at 0:45
user3203476
asked Nov 13 '15 at 0:45
user3203476user3203476
746614
746614
$begingroup$
Is it clear to you that the betting strategy achieves what the author claims? That is, that if our gambler throws a T on the $n^{th}$ turn (prior to a win) that the net position is $-n$?
$endgroup$
– lulu
Nov 13 '15 at 1:28
$begingroup$
Sorry no it is not :(
$endgroup$
– user3203476
Nov 13 '15 at 1:29
$begingroup$
Ok. Write out several cases (avoiding the string $HHH$). Confirm that it works. Then proceed by induction. The strategy really is the whole point. The bets are specifically structured to ensure that payout.
$endgroup$
– lulu
Nov 13 '15 at 1:30
$begingroup$
The logic of the proof is: "we've given the rules for a betting strategy. From general theory, it is not possible that these rules (nor any other set of rules) has anything other that a $0$ expectation. The beauty of this strategy is that we can actually computed the payout after the expected number of turns.
$endgroup$
– lulu
Nov 13 '15 at 1:33
$begingroup$
This is a really neat way to solve this problem. Consider a smaller scenario - expected time to get one $H$. Bet on $H$ so that if you get $T$ on your $nth$ throw, your position is $-n$. Stop if you get an $H$. In other words, always bet one. If you get an $H$ on the $nth$ throw, your position will be $1-(n-1)=2-n$ and hence $E(n)=2$.
$endgroup$
– A.S.
Nov 13 '15 at 3:52
|
show 1 more comment
$begingroup$
Is it clear to you that the betting strategy achieves what the author claims? That is, that if our gambler throws a T on the $n^{th}$ turn (prior to a win) that the net position is $-n$?
$endgroup$
– lulu
Nov 13 '15 at 1:28
$begingroup$
Sorry no it is not :(
$endgroup$
– user3203476
Nov 13 '15 at 1:29
$begingroup$
Ok. Write out several cases (avoiding the string $HHH$). Confirm that it works. Then proceed by induction. The strategy really is the whole point. The bets are specifically structured to ensure that payout.
$endgroup$
– lulu
Nov 13 '15 at 1:30
$begingroup$
The logic of the proof is: "we've given the rules for a betting strategy. From general theory, it is not possible that these rules (nor any other set of rules) has anything other that a $0$ expectation. The beauty of this strategy is that we can actually computed the payout after the expected number of turns.
$endgroup$
– lulu
Nov 13 '15 at 1:33
$begingroup$
This is a really neat way to solve this problem. Consider a smaller scenario - expected time to get one $H$. Bet on $H$ so that if you get $T$ on your $nth$ throw, your position is $-n$. Stop if you get an $H$. In other words, always bet one. If you get an $H$ on the $nth$ throw, your position will be $1-(n-1)=2-n$ and hence $E(n)=2$.
$endgroup$
– A.S.
Nov 13 '15 at 3:52
$begingroup$
Is it clear to you that the betting strategy achieves what the author claims? That is, that if our gambler throws a T on the $n^{th}$ turn (prior to a win) that the net position is $-n$?
$endgroup$
– lulu
Nov 13 '15 at 1:28
$begingroup$
Is it clear to you that the betting strategy achieves what the author claims? That is, that if our gambler throws a T on the $n^{th}$ turn (prior to a win) that the net position is $-n$?
$endgroup$
– lulu
Nov 13 '15 at 1:28
$begingroup$
Sorry no it is not :(
$endgroup$
– user3203476
Nov 13 '15 at 1:29
$begingroup$
Sorry no it is not :(
$endgroup$
– user3203476
Nov 13 '15 at 1:29
$begingroup$
Ok. Write out several cases (avoiding the string $HHH$). Confirm that it works. Then proceed by induction. The strategy really is the whole point. The bets are specifically structured to ensure that payout.
$endgroup$
– lulu
Nov 13 '15 at 1:30
$begingroup$
Ok. Write out several cases (avoiding the string $HHH$). Confirm that it works. Then proceed by induction. The strategy really is the whole point. The bets are specifically structured to ensure that payout.
$endgroup$
– lulu
Nov 13 '15 at 1:30
$begingroup$
The logic of the proof is: "we've given the rules for a betting strategy. From general theory, it is not possible that these rules (nor any other set of rules) has anything other that a $0$ expectation. The beauty of this strategy is that we can actually computed the payout after the expected number of turns.
$endgroup$
– lulu
Nov 13 '15 at 1:33
$begingroup$
The logic of the proof is: "we've given the rules for a betting strategy. From general theory, it is not possible that these rules (nor any other set of rules) has anything other that a $0$ expectation. The beauty of this strategy is that we can actually computed the payout after the expected number of turns.
$endgroup$
– lulu
Nov 13 '15 at 1:33
$begingroup$
This is a really neat way to solve this problem. Consider a smaller scenario - expected time to get one $H$. Bet on $H$ so that if you get $T$ on your $nth$ throw, your position is $-n$. Stop if you get an $H$. In other words, always bet one. If you get an $H$ on the $nth$ throw, your position will be $1-(n-1)=2-n$ and hence $E(n)=2$.
$endgroup$
– A.S.
Nov 13 '15 at 3:52
$begingroup$
This is a really neat way to solve this problem. Consider a smaller scenario - expected time to get one $H$. Bet on $H$ so that if you get $T$ on your $nth$ throw, your position is $-n$. Stop if you get an $H$. In other words, always bet one. If you get an $H$ on the $nth$ throw, your position will be $1-(n-1)=2-n$ and hence $E(n)=2$.
$endgroup$
– A.S.
Nov 13 '15 at 3:52
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
To be honest, I didn't really enjoy the explanations of the author... I understand that he wants to motivate the use of martingale but it over complicates a simple problem.
Let $E$ be the expected #of tosses to achieve 3 heads, if we throw for example $HT$ then we have to start again counting, so in the particular case $E$ will be augmented by 2; enumerates all cases (HHH, HHT, HT, T) and weight by their probability
$$ E=frac{1}{8}3+frac{1}{8}(3+E)+frac{1}{4}(2+E)+frac{1}{2}(1+E)$$
Solve for E.
I understand there is no martingale argument in that reasoning, but I also see little value in forcing concepts where there aren't needed.
$endgroup$
add a comment |
$begingroup$
I believe that the strategy is that:
consider the payoff of the $i_{th}$ toss
(1) If $i == 1$ or $X_{i-1} = Tail$ , you bet 1 unit on Head. Which means that $Payoff_i(X_i=Tail) = -1$ $Payoff_i(X_i=Head) = 1$;
(2) If $X_{i-1} = Head$ , you bet 3 unit on Head. Which means that $Payoff_i(X_i=Tail) = -3$ $Payoff_i(X_i=Head) = 3$;
(3) If $X_{i-2} = Head$ and $X_{i-1} = Head$ , you bet 7 unit on Head. Which means that $Payoff_i(X_i=Tail) = -7$ $Payoff_i(X_i=Head) = 7$;
(4) stop tossing when you get HHH serial.
Conclusions about this strategy:
(1) If you get a Tail at $i_{th}$ toss, your total payoff is $-n$(you could test this conclusion for arbitrage simple series);
(2)For every toss, your expected payoff is zero. Your total payoff is a martingale.
$$ text{
**any bounded trading strategy in a martingale is a martingale**
}
$$
I think that the first "martingale" in this quote is the total payoff $M_t = sum_{s=0}^tpayoff_s$.
The trading strategy descripted above is also a maringale.
When this strategy stop, denote $n$ as the strategy stopping time. The expected payoff should be zero.which is
$E[-(n-4)+1+3+7]=14-E[n]=0$. $E[n]=14$.
$endgroup$
add a comment |
$begingroup$
I believe the author is using a great deal of words to define the wealth process : $$M_n = M_0 + sum_{1}^{3} 2^{i}X_{i}text{ with }M_0=1text{ and } X_i text{ iid as }{1,-1}text{ with p=}frac{1}{2}$$
$mathbb E(M_n|M_{n-1}) = mathbb E(M_{n-1}+2^nX_n|M_{n-1})=M_{n-1}+2^nmathbb E(X_{n-1})=M_{n-1}$ so $M_n$ is a martingale
then for the stopping time $tau = tau_{HHH}+1$ since we want the number of throws rather than the number of $M_i$s
$$mathbb E (M_{tau}) = 1 + 2^1 + 2^2 + 2^3 = mathbb E(M_0)mathbb E(tau)=mathbb E_{tau}=1+E(tau_{HHH})text{ so } mathbb E(tau_{HHH})=14$$
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This doesn't seem like the process the author defined. What are the bounds of summation in definition of $M_n$? How is $E(M_tau)=E(M_0)E(tau)$?
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– A.S.
Nov 15 '15 at 9:13
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you are right the Wald identity does not apply. I will amend or delete if I cannot find anything elegant. I would like to avoid this 'team of gamblers' method one usually sees in this context.
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– user3203476
Nov 15 '15 at 12:56
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He starts with $M_0=0$, sets $M_n=M_{n-1}+X_n(n+M_{n-1})$ - a martingale.
$endgroup$
– A.S.
Nov 15 '15 at 22:07
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To be honest, I didn't really enjoy the explanations of the author... I understand that he wants to motivate the use of martingale but it over complicates a simple problem.
Let $E$ be the expected #of tosses to achieve 3 heads, if we throw for example $HT$ then we have to start again counting, so in the particular case $E$ will be augmented by 2; enumerates all cases (HHH, HHT, HT, T) and weight by their probability
$$ E=frac{1}{8}3+frac{1}{8}(3+E)+frac{1}{4}(2+E)+frac{1}{2}(1+E)$$
Solve for E.
I understand there is no martingale argument in that reasoning, but I also see little value in forcing concepts where there aren't needed.
$endgroup$
add a comment |
$begingroup$
To be honest, I didn't really enjoy the explanations of the author... I understand that he wants to motivate the use of martingale but it over complicates a simple problem.
Let $E$ be the expected #of tosses to achieve 3 heads, if we throw for example $HT$ then we have to start again counting, so in the particular case $E$ will be augmented by 2; enumerates all cases (HHH, HHT, HT, T) and weight by their probability
$$ E=frac{1}{8}3+frac{1}{8}(3+E)+frac{1}{4}(2+E)+frac{1}{2}(1+E)$$
Solve for E.
I understand there is no martingale argument in that reasoning, but I also see little value in forcing concepts where there aren't needed.
$endgroup$
add a comment |
$begingroup$
To be honest, I didn't really enjoy the explanations of the author... I understand that he wants to motivate the use of martingale but it over complicates a simple problem.
Let $E$ be the expected #of tosses to achieve 3 heads, if we throw for example $HT$ then we have to start again counting, so in the particular case $E$ will be augmented by 2; enumerates all cases (HHH, HHT, HT, T) and weight by their probability
$$ E=frac{1}{8}3+frac{1}{8}(3+E)+frac{1}{4}(2+E)+frac{1}{2}(1+E)$$
Solve for E.
I understand there is no martingale argument in that reasoning, but I also see little value in forcing concepts where there aren't needed.
$endgroup$
To be honest, I didn't really enjoy the explanations of the author... I understand that he wants to motivate the use of martingale but it over complicates a simple problem.
Let $E$ be the expected #of tosses to achieve 3 heads, if we throw for example $HT$ then we have to start again counting, so in the particular case $E$ will be augmented by 2; enumerates all cases (HHH, HHT, HT, T) and weight by their probability
$$ E=frac{1}{8}3+frac{1}{8}(3+E)+frac{1}{4}(2+E)+frac{1}{2}(1+E)$$
Solve for E.
I understand there is no martingale argument in that reasoning, but I also see little value in forcing concepts where there aren't needed.
answered Nov 13 '15 at 2:56
zebullonzebullon
397212
397212
add a comment |
add a comment |
$begingroup$
I believe that the strategy is that:
consider the payoff of the $i_{th}$ toss
(1) If $i == 1$ or $X_{i-1} = Tail$ , you bet 1 unit on Head. Which means that $Payoff_i(X_i=Tail) = -1$ $Payoff_i(X_i=Head) = 1$;
(2) If $X_{i-1} = Head$ , you bet 3 unit on Head. Which means that $Payoff_i(X_i=Tail) = -3$ $Payoff_i(X_i=Head) = 3$;
(3) If $X_{i-2} = Head$ and $X_{i-1} = Head$ , you bet 7 unit on Head. Which means that $Payoff_i(X_i=Tail) = -7$ $Payoff_i(X_i=Head) = 7$;
(4) stop tossing when you get HHH serial.
Conclusions about this strategy:
(1) If you get a Tail at $i_{th}$ toss, your total payoff is $-n$(you could test this conclusion for arbitrage simple series);
(2)For every toss, your expected payoff is zero. Your total payoff is a martingale.
$$ text{
**any bounded trading strategy in a martingale is a martingale**
}
$$
I think that the first "martingale" in this quote is the total payoff $M_t = sum_{s=0}^tpayoff_s$.
The trading strategy descripted above is also a maringale.
When this strategy stop, denote $n$ as the strategy stopping time. The expected payoff should be zero.which is
$E[-(n-4)+1+3+7]=14-E[n]=0$. $E[n]=14$.
$endgroup$
add a comment |
$begingroup$
I believe that the strategy is that:
consider the payoff of the $i_{th}$ toss
(1) If $i == 1$ or $X_{i-1} = Tail$ , you bet 1 unit on Head. Which means that $Payoff_i(X_i=Tail) = -1$ $Payoff_i(X_i=Head) = 1$;
(2) If $X_{i-1} = Head$ , you bet 3 unit on Head. Which means that $Payoff_i(X_i=Tail) = -3$ $Payoff_i(X_i=Head) = 3$;
(3) If $X_{i-2} = Head$ and $X_{i-1} = Head$ , you bet 7 unit on Head. Which means that $Payoff_i(X_i=Tail) = -7$ $Payoff_i(X_i=Head) = 7$;
(4) stop tossing when you get HHH serial.
Conclusions about this strategy:
(1) If you get a Tail at $i_{th}$ toss, your total payoff is $-n$(you could test this conclusion for arbitrage simple series);
(2)For every toss, your expected payoff is zero. Your total payoff is a martingale.
$$ text{
**any bounded trading strategy in a martingale is a martingale**
}
$$
I think that the first "martingale" in this quote is the total payoff $M_t = sum_{s=0}^tpayoff_s$.
The trading strategy descripted above is also a maringale.
When this strategy stop, denote $n$ as the strategy stopping time. The expected payoff should be zero.which is
$E[-(n-4)+1+3+7]=14-E[n]=0$. $E[n]=14$.
$endgroup$
add a comment |
$begingroup$
I believe that the strategy is that:
consider the payoff of the $i_{th}$ toss
(1) If $i == 1$ or $X_{i-1} = Tail$ , you bet 1 unit on Head. Which means that $Payoff_i(X_i=Tail) = -1$ $Payoff_i(X_i=Head) = 1$;
(2) If $X_{i-1} = Head$ , you bet 3 unit on Head. Which means that $Payoff_i(X_i=Tail) = -3$ $Payoff_i(X_i=Head) = 3$;
(3) If $X_{i-2} = Head$ and $X_{i-1} = Head$ , you bet 7 unit on Head. Which means that $Payoff_i(X_i=Tail) = -7$ $Payoff_i(X_i=Head) = 7$;
(4) stop tossing when you get HHH serial.
Conclusions about this strategy:
(1) If you get a Tail at $i_{th}$ toss, your total payoff is $-n$(you could test this conclusion for arbitrage simple series);
(2)For every toss, your expected payoff is zero. Your total payoff is a martingale.
$$ text{
**any bounded trading strategy in a martingale is a martingale**
}
$$
I think that the first "martingale" in this quote is the total payoff $M_t = sum_{s=0}^tpayoff_s$.
The trading strategy descripted above is also a maringale.
When this strategy stop, denote $n$ as the strategy stopping time. The expected payoff should be zero.which is
$E[-(n-4)+1+3+7]=14-E[n]=0$. $E[n]=14$.
$endgroup$
I believe that the strategy is that:
consider the payoff of the $i_{th}$ toss
(1) If $i == 1$ or $X_{i-1} = Tail$ , you bet 1 unit on Head. Which means that $Payoff_i(X_i=Tail) = -1$ $Payoff_i(X_i=Head) = 1$;
(2) If $X_{i-1} = Head$ , you bet 3 unit on Head. Which means that $Payoff_i(X_i=Tail) = -3$ $Payoff_i(X_i=Head) = 3$;
(3) If $X_{i-2} = Head$ and $X_{i-1} = Head$ , you bet 7 unit on Head. Which means that $Payoff_i(X_i=Tail) = -7$ $Payoff_i(X_i=Head) = 7$;
(4) stop tossing when you get HHH serial.
Conclusions about this strategy:
(1) If you get a Tail at $i_{th}$ toss, your total payoff is $-n$(you could test this conclusion for arbitrage simple series);
(2)For every toss, your expected payoff is zero. Your total payoff is a martingale.
$$ text{
**any bounded trading strategy in a martingale is a martingale**
}
$$
I think that the first "martingale" in this quote is the total payoff $M_t = sum_{s=0}^tpayoff_s$.
The trading strategy descripted above is also a maringale.
When this strategy stop, denote $n$ as the strategy stopping time. The expected payoff should be zero.which is
$E[-(n-4)+1+3+7]=14-E[n]=0$. $E[n]=14$.
answered Dec 6 '17 at 11:33
LexiLexi
63
63
add a comment |
add a comment |
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I believe the author is using a great deal of words to define the wealth process : $$M_n = M_0 + sum_{1}^{3} 2^{i}X_{i}text{ with }M_0=1text{ and } X_i text{ iid as }{1,-1}text{ with p=}frac{1}{2}$$
$mathbb E(M_n|M_{n-1}) = mathbb E(M_{n-1}+2^nX_n|M_{n-1})=M_{n-1}+2^nmathbb E(X_{n-1})=M_{n-1}$ so $M_n$ is a martingale
then for the stopping time $tau = tau_{HHH}+1$ since we want the number of throws rather than the number of $M_i$s
$$mathbb E (M_{tau}) = 1 + 2^1 + 2^2 + 2^3 = mathbb E(M_0)mathbb E(tau)=mathbb E_{tau}=1+E(tau_{HHH})text{ so } mathbb E(tau_{HHH})=14$$
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This doesn't seem like the process the author defined. What are the bounds of summation in definition of $M_n$? How is $E(M_tau)=E(M_0)E(tau)$?
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– A.S.
Nov 15 '15 at 9:13
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you are right the Wald identity does not apply. I will amend or delete if I cannot find anything elegant. I would like to avoid this 'team of gamblers' method one usually sees in this context.
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– user3203476
Nov 15 '15 at 12:56
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He starts with $M_0=0$, sets $M_n=M_{n-1}+X_n(n+M_{n-1})$ - a martingale.
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– A.S.
Nov 15 '15 at 22:07
add a comment |
$begingroup$
I believe the author is using a great deal of words to define the wealth process : $$M_n = M_0 + sum_{1}^{3} 2^{i}X_{i}text{ with }M_0=1text{ and } X_i text{ iid as }{1,-1}text{ with p=}frac{1}{2}$$
$mathbb E(M_n|M_{n-1}) = mathbb E(M_{n-1}+2^nX_n|M_{n-1})=M_{n-1}+2^nmathbb E(X_{n-1})=M_{n-1}$ so $M_n$ is a martingale
then for the stopping time $tau = tau_{HHH}+1$ since we want the number of throws rather than the number of $M_i$s
$$mathbb E (M_{tau}) = 1 + 2^1 + 2^2 + 2^3 = mathbb E(M_0)mathbb E(tau)=mathbb E_{tau}=1+E(tau_{HHH})text{ so } mathbb E(tau_{HHH})=14$$
$endgroup$
$begingroup$
This doesn't seem like the process the author defined. What are the bounds of summation in definition of $M_n$? How is $E(M_tau)=E(M_0)E(tau)$?
$endgroup$
– A.S.
Nov 15 '15 at 9:13
$begingroup$
you are right the Wald identity does not apply. I will amend or delete if I cannot find anything elegant. I would like to avoid this 'team of gamblers' method one usually sees in this context.
$endgroup$
– user3203476
Nov 15 '15 at 12:56
$begingroup$
He starts with $M_0=0$, sets $M_n=M_{n-1}+X_n(n+M_{n-1})$ - a martingale.
$endgroup$
– A.S.
Nov 15 '15 at 22:07
add a comment |
$begingroup$
I believe the author is using a great deal of words to define the wealth process : $$M_n = M_0 + sum_{1}^{3} 2^{i}X_{i}text{ with }M_0=1text{ and } X_i text{ iid as }{1,-1}text{ with p=}frac{1}{2}$$
$mathbb E(M_n|M_{n-1}) = mathbb E(M_{n-1}+2^nX_n|M_{n-1})=M_{n-1}+2^nmathbb E(X_{n-1})=M_{n-1}$ so $M_n$ is a martingale
then for the stopping time $tau = tau_{HHH}+1$ since we want the number of throws rather than the number of $M_i$s
$$mathbb E (M_{tau}) = 1 + 2^1 + 2^2 + 2^3 = mathbb E(M_0)mathbb E(tau)=mathbb E_{tau}=1+E(tau_{HHH})text{ so } mathbb E(tau_{HHH})=14$$
$endgroup$
I believe the author is using a great deal of words to define the wealth process : $$M_n = M_0 + sum_{1}^{3} 2^{i}X_{i}text{ with }M_0=1text{ and } X_i text{ iid as }{1,-1}text{ with p=}frac{1}{2}$$
$mathbb E(M_n|M_{n-1}) = mathbb E(M_{n-1}+2^nX_n|M_{n-1})=M_{n-1}+2^nmathbb E(X_{n-1})=M_{n-1}$ so $M_n$ is a martingale
then for the stopping time $tau = tau_{HHH}+1$ since we want the number of throws rather than the number of $M_i$s
$$mathbb E (M_{tau}) = 1 + 2^1 + 2^2 + 2^3 = mathbb E(M_0)mathbb E(tau)=mathbb E_{tau}=1+E(tau_{HHH})text{ so } mathbb E(tau_{HHH})=14$$
edited Nov 15 '15 at 8:47
answered Nov 15 '15 at 7:07
user3203476user3203476
746614
746614
$begingroup$
This doesn't seem like the process the author defined. What are the bounds of summation in definition of $M_n$? How is $E(M_tau)=E(M_0)E(tau)$?
$endgroup$
– A.S.
Nov 15 '15 at 9:13
$begingroup$
you are right the Wald identity does not apply. I will amend or delete if I cannot find anything elegant. I would like to avoid this 'team of gamblers' method one usually sees in this context.
$endgroup$
– user3203476
Nov 15 '15 at 12:56
$begingroup$
He starts with $M_0=0$, sets $M_n=M_{n-1}+X_n(n+M_{n-1})$ - a martingale.
$endgroup$
– A.S.
Nov 15 '15 at 22:07
add a comment |
$begingroup$
This doesn't seem like the process the author defined. What are the bounds of summation in definition of $M_n$? How is $E(M_tau)=E(M_0)E(tau)$?
$endgroup$
– A.S.
Nov 15 '15 at 9:13
$begingroup$
you are right the Wald identity does not apply. I will amend or delete if I cannot find anything elegant. I would like to avoid this 'team of gamblers' method one usually sees in this context.
$endgroup$
– user3203476
Nov 15 '15 at 12:56
$begingroup$
He starts with $M_0=0$, sets $M_n=M_{n-1}+X_n(n+M_{n-1})$ - a martingale.
$endgroup$
– A.S.
Nov 15 '15 at 22:07
$begingroup$
This doesn't seem like the process the author defined. What are the bounds of summation in definition of $M_n$? How is $E(M_tau)=E(M_0)E(tau)$?
$endgroup$
– A.S.
Nov 15 '15 at 9:13
$begingroup$
This doesn't seem like the process the author defined. What are the bounds of summation in definition of $M_n$? How is $E(M_tau)=E(M_0)E(tau)$?
$endgroup$
– A.S.
Nov 15 '15 at 9:13
$begingroup$
you are right the Wald identity does not apply. I will amend or delete if I cannot find anything elegant. I would like to avoid this 'team of gamblers' method one usually sees in this context.
$endgroup$
– user3203476
Nov 15 '15 at 12:56
$begingroup$
you are right the Wald identity does not apply. I will amend or delete if I cannot find anything elegant. I would like to avoid this 'team of gamblers' method one usually sees in this context.
$endgroup$
– user3203476
Nov 15 '15 at 12:56
$begingroup$
He starts with $M_0=0$, sets $M_n=M_{n-1}+X_n(n+M_{n-1})$ - a martingale.
$endgroup$
– A.S.
Nov 15 '15 at 22:07
$begingroup$
He starts with $M_0=0$, sets $M_n=M_{n-1}+X_n(n+M_{n-1})$ - a martingale.
$endgroup$
– A.S.
Nov 15 '15 at 22:07
add a comment |
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$begingroup$
Is it clear to you that the betting strategy achieves what the author claims? That is, that if our gambler throws a T on the $n^{th}$ turn (prior to a win) that the net position is $-n$?
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– lulu
Nov 13 '15 at 1:28
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Sorry no it is not :(
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– user3203476
Nov 13 '15 at 1:29
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Ok. Write out several cases (avoiding the string $HHH$). Confirm that it works. Then proceed by induction. The strategy really is the whole point. The bets are specifically structured to ensure that payout.
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– lulu
Nov 13 '15 at 1:30
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The logic of the proof is: "we've given the rules for a betting strategy. From general theory, it is not possible that these rules (nor any other set of rules) has anything other that a $0$ expectation. The beauty of this strategy is that we can actually computed the payout after the expected number of turns.
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– lulu
Nov 13 '15 at 1:33
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This is a really neat way to solve this problem. Consider a smaller scenario - expected time to get one $H$. Bet on $H$ so that if you get $T$ on your $nth$ throw, your position is $-n$. Stop if you get an $H$. In other words, always bet one. If you get an $H$ on the $nth$ throw, your position will be $1-(n-1)=2-n$ and hence $E(n)=2$.
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– A.S.
Nov 13 '15 at 3:52