Showing that a market model has arbitrage and describing martingales
This is an exercise which I came upon while studying an introduction to financial mathematics.
Exercise :
Consider the finite sample space $Omega = {omega_1,omega_2,omega_3}$ and let $mathbb P$ be a probability measure such that $mathbb P[{omega_1}] > 0$ for all $i=1,2,3$. We define a financial market of one period which is consisted by the probability space $(Omega,mathcal{F},mathbb P)$ with $mathcal{F} := 2^Omega$ and the securities $bar{S} = (S^0,S^1,S^2)$ which are consisted of the zero-risk security $S^0$ and two securities $S^1,S^2$ which have risk. Their values at the time $t=0$ are given by the vector
$$bar{S}_0 = begin{pmatrix} 1\2\7 end{pmatrix}$$
while their values at time $t=1$, depending whether the scenario $omega_1,omega_2$ or $omega_3$ happens, are given by the vectors
$$bar{S}_1(omega_1) = begin{pmatrix} 1\3\9end{pmatrix}, quad bar{S}_1(omega_2) = begin{pmatrix} 1\1\5end{pmatrix}, quad bar{S}_1(omega_3) = begin{pmatrix} 1\5\10 end{pmatrix}$$
(a) Show that this financial market has arbitrage.
(b) Let $S_1^2(omega_3) = 13$ while the other values remain the same as before. Show that this financial market does not have arbitrage and describe all the equivalent martingale measures.
Attempt :
(a) We have that a value process is defined as :
$$V_t = V_t^bar{xi} = bar{xi}cdot bar{S}_t = sum_{i=0}^d xi_t^icdot bar{S}_t^i, quad t in {0,1}$$
where $xi = (xi^0, xi) in mathbb R^{d+1}$ is an investment strategy where the number $xi^i$ is equal to the number of pieces from the security $S^i$ which are contained in the portfolio at the time period $[0,1], i in {0,1,dots,d}$.
Now, I also know that to show that a market has arbitrage, I need to show the following :
$$V_0 leq 0, quad mathbb P(V1 geq 0) = 1, quad mathbb P(V_1 > 0) > 0$$
I understand that the different $S$ vectors will be plugged in to calculate $V_t$ but I really can't comprehend $xi$. What would the $xi$ vector be ?
Any help for me to understand what $xi$ really is based on the problem and how to complete my attempt will be much appreciated.
For (b), showing that it does not have arbitrage is similar to (a) as I will just show that one of these conditions will not hold. What about the martingale stuff though ? It's a mathematical substance we really haven't been into so, if possible, I would really appreciate an elaborations.
stochastic-processes arbitrage finance-mathematics martingale no-arbitrage-theory
add a comment |
This is an exercise which I came upon while studying an introduction to financial mathematics.
Exercise :
Consider the finite sample space $Omega = {omega_1,omega_2,omega_3}$ and let $mathbb P$ be a probability measure such that $mathbb P[{omega_1}] > 0$ for all $i=1,2,3$. We define a financial market of one period which is consisted by the probability space $(Omega,mathcal{F},mathbb P)$ with $mathcal{F} := 2^Omega$ and the securities $bar{S} = (S^0,S^1,S^2)$ which are consisted of the zero-risk security $S^0$ and two securities $S^1,S^2$ which have risk. Their values at the time $t=0$ are given by the vector
$$bar{S}_0 = begin{pmatrix} 1\2\7 end{pmatrix}$$
while their values at time $t=1$, depending whether the scenario $omega_1,omega_2$ or $omega_3$ happens, are given by the vectors
$$bar{S}_1(omega_1) = begin{pmatrix} 1\3\9end{pmatrix}, quad bar{S}_1(omega_2) = begin{pmatrix} 1\1\5end{pmatrix}, quad bar{S}_1(omega_3) = begin{pmatrix} 1\5\10 end{pmatrix}$$
(a) Show that this financial market has arbitrage.
(b) Let $S_1^2(omega_3) = 13$ while the other values remain the same as before. Show that this financial market does not have arbitrage and describe all the equivalent martingale measures.
Attempt :
(a) We have that a value process is defined as :
$$V_t = V_t^bar{xi} = bar{xi}cdot bar{S}_t = sum_{i=0}^d xi_t^icdot bar{S}_t^i, quad t in {0,1}$$
where $xi = (xi^0, xi) in mathbb R^{d+1}$ is an investment strategy where the number $xi^i$ is equal to the number of pieces from the security $S^i$ which are contained in the portfolio at the time period $[0,1], i in {0,1,dots,d}$.
Now, I also know that to show that a market has arbitrage, I need to show the following :
$$V_0 leq 0, quad mathbb P(V1 geq 0) = 1, quad mathbb P(V_1 > 0) > 0$$
I understand that the different $S$ vectors will be plugged in to calculate $V_t$ but I really can't comprehend $xi$. What would the $xi$ vector be ?
Any help for me to understand what $xi$ really is based on the problem and how to complete my attempt will be much appreciated.
For (b), showing that it does not have arbitrage is similar to (a) as I will just show that one of these conditions will not hold. What about the martingale stuff though ? It's a mathematical substance we really haven't been into so, if possible, I would really appreciate an elaborations.
stochastic-processes arbitrage finance-mathematics martingale no-arbitrage-theory
add a comment |
This is an exercise which I came upon while studying an introduction to financial mathematics.
Exercise :
Consider the finite sample space $Omega = {omega_1,omega_2,omega_3}$ and let $mathbb P$ be a probability measure such that $mathbb P[{omega_1}] > 0$ for all $i=1,2,3$. We define a financial market of one period which is consisted by the probability space $(Omega,mathcal{F},mathbb P)$ with $mathcal{F} := 2^Omega$ and the securities $bar{S} = (S^0,S^1,S^2)$ which are consisted of the zero-risk security $S^0$ and two securities $S^1,S^2$ which have risk. Their values at the time $t=0$ are given by the vector
$$bar{S}_0 = begin{pmatrix} 1\2\7 end{pmatrix}$$
while their values at time $t=1$, depending whether the scenario $omega_1,omega_2$ or $omega_3$ happens, are given by the vectors
$$bar{S}_1(omega_1) = begin{pmatrix} 1\3\9end{pmatrix}, quad bar{S}_1(omega_2) = begin{pmatrix} 1\1\5end{pmatrix}, quad bar{S}_1(omega_3) = begin{pmatrix} 1\5\10 end{pmatrix}$$
(a) Show that this financial market has arbitrage.
(b) Let $S_1^2(omega_3) = 13$ while the other values remain the same as before. Show that this financial market does not have arbitrage and describe all the equivalent martingale measures.
Attempt :
(a) We have that a value process is defined as :
$$V_t = V_t^bar{xi} = bar{xi}cdot bar{S}_t = sum_{i=0}^d xi_t^icdot bar{S}_t^i, quad t in {0,1}$$
where $xi = (xi^0, xi) in mathbb R^{d+1}$ is an investment strategy where the number $xi^i$ is equal to the number of pieces from the security $S^i$ which are contained in the portfolio at the time period $[0,1], i in {0,1,dots,d}$.
Now, I also know that to show that a market has arbitrage, I need to show the following :
$$V_0 leq 0, quad mathbb P(V1 geq 0) = 1, quad mathbb P(V_1 > 0) > 0$$
I understand that the different $S$ vectors will be plugged in to calculate $V_t$ but I really can't comprehend $xi$. What would the $xi$ vector be ?
Any help for me to understand what $xi$ really is based on the problem and how to complete my attempt will be much appreciated.
For (b), showing that it does not have arbitrage is similar to (a) as I will just show that one of these conditions will not hold. What about the martingale stuff though ? It's a mathematical substance we really haven't been into so, if possible, I would really appreciate an elaborations.
stochastic-processes arbitrage finance-mathematics martingale no-arbitrage-theory
This is an exercise which I came upon while studying an introduction to financial mathematics.
Exercise :
Consider the finite sample space $Omega = {omega_1,omega_2,omega_3}$ and let $mathbb P$ be a probability measure such that $mathbb P[{omega_1}] > 0$ for all $i=1,2,3$. We define a financial market of one period which is consisted by the probability space $(Omega,mathcal{F},mathbb P)$ with $mathcal{F} := 2^Omega$ and the securities $bar{S} = (S^0,S^1,S^2)$ which are consisted of the zero-risk security $S^0$ and two securities $S^1,S^2$ which have risk. Their values at the time $t=0$ are given by the vector
$$bar{S}_0 = begin{pmatrix} 1\2\7 end{pmatrix}$$
while their values at time $t=1$, depending whether the scenario $omega_1,omega_2$ or $omega_3$ happens, are given by the vectors
$$bar{S}_1(omega_1) = begin{pmatrix} 1\3\9end{pmatrix}, quad bar{S}_1(omega_2) = begin{pmatrix} 1\1\5end{pmatrix}, quad bar{S}_1(omega_3) = begin{pmatrix} 1\5\10 end{pmatrix}$$
(a) Show that this financial market has arbitrage.
(b) Let $S_1^2(omega_3) = 13$ while the other values remain the same as before. Show that this financial market does not have arbitrage and describe all the equivalent martingale measures.
Attempt :
(a) We have that a value process is defined as :
$$V_t = V_t^bar{xi} = bar{xi}cdot bar{S}_t = sum_{i=0}^d xi_t^icdot bar{S}_t^i, quad t in {0,1}$$
where $xi = (xi^0, xi) in mathbb R^{d+1}$ is an investment strategy where the number $xi^i$ is equal to the number of pieces from the security $S^i$ which are contained in the portfolio at the time period $[0,1], i in {0,1,dots,d}$.
Now, I also know that to show that a market has arbitrage, I need to show the following :
$$V_0 leq 0, quad mathbb P(V1 geq 0) = 1, quad mathbb P(V_1 > 0) > 0$$
I understand that the different $S$ vectors will be plugged in to calculate $V_t$ but I really can't comprehend $xi$. What would the $xi$ vector be ?
Any help for me to understand what $xi$ really is based on the problem and how to complete my attempt will be much appreciated.
For (b), showing that it does not have arbitrage is similar to (a) as I will just show that one of these conditions will not hold. What about the martingale stuff though ? It's a mathematical substance we really haven't been into so, if possible, I would really appreciate an elaborations.
stochastic-processes arbitrage finance-mathematics martingale no-arbitrage-theory
stochastic-processes arbitrage finance-mathematics martingale no-arbitrage-theory
asked Nov 11 at 18:10
Rebellos
1284
1284
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1 Answer
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The parameter $xi$ represents your strategy, namely the quantity you hold in your portfolio of each security $S^0$, $S^1$ and $S^2$. Consider the following strategy:
$${xi}=(xi^1,xi^2,xi^3)=(1.5,1,-0.5)$$
Then:
$$begin{align}
& t=0: && xibar{S}_0=xi^0S_0^0+xi^1S_0^1+xi^2S_0^2 = 1.5+2-3.5=0
\
& t=1: && xibar{S}_1(omega_1)=1.5+3-4.5=0
\
&&& xibar{S}_1(omega_2)=1.5+1-2.5=0
\
&&& xibar{S}_1(omega_3)=1.5+5-5=1.5>0
end{align}$$
Thus:
$$xibar{S}_0=0, quad mathbb{P}(xibar{S}_1geq0)=1, quad mathbb{P}(xibar{S}_1>0)>0$$
Hence the market has arbitrage.
For question b), you need to generalize to prove that there is no portfolio $xi$ that allows arbitrage (instead of just finding a counterexample as in a).
Hello, why is it legit to just take a random $xi$ and show the conditions ? Wouldn't the $xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ?
– Rebellos
Nov 11 at 20:51
It is not a random $xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $xi$, and there is a logic to it.
– Alex C
Nov 11 at 22:48
@Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(omega_1)/S_1^2(omega_1), S_1^1(omega_3)/S_1^2(omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(omega_2)/S_1^1(omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$.
– Daneel Olivaw
Nov 11 at 23:02
Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do?
– Rebellos
Nov 11 at 23:06
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
The parameter $xi$ represents your strategy, namely the quantity you hold in your portfolio of each security $S^0$, $S^1$ and $S^2$. Consider the following strategy:
$${xi}=(xi^1,xi^2,xi^3)=(1.5,1,-0.5)$$
Then:
$$begin{align}
& t=0: && xibar{S}_0=xi^0S_0^0+xi^1S_0^1+xi^2S_0^2 = 1.5+2-3.5=0
\
& t=1: && xibar{S}_1(omega_1)=1.5+3-4.5=0
\
&&& xibar{S}_1(omega_2)=1.5+1-2.5=0
\
&&& xibar{S}_1(omega_3)=1.5+5-5=1.5>0
end{align}$$
Thus:
$$xibar{S}_0=0, quad mathbb{P}(xibar{S}_1geq0)=1, quad mathbb{P}(xibar{S}_1>0)>0$$
Hence the market has arbitrage.
For question b), you need to generalize to prove that there is no portfolio $xi$ that allows arbitrage (instead of just finding a counterexample as in a).
Hello, why is it legit to just take a random $xi$ and show the conditions ? Wouldn't the $xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ?
– Rebellos
Nov 11 at 20:51
It is not a random $xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $xi$, and there is a logic to it.
– Alex C
Nov 11 at 22:48
@Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(omega_1)/S_1^2(omega_1), S_1^1(omega_3)/S_1^2(omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(omega_2)/S_1^1(omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$.
– Daneel Olivaw
Nov 11 at 23:02
Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do?
– Rebellos
Nov 11 at 23:06
add a comment |
The parameter $xi$ represents your strategy, namely the quantity you hold in your portfolio of each security $S^0$, $S^1$ and $S^2$. Consider the following strategy:
$${xi}=(xi^1,xi^2,xi^3)=(1.5,1,-0.5)$$
Then:
$$begin{align}
& t=0: && xibar{S}_0=xi^0S_0^0+xi^1S_0^1+xi^2S_0^2 = 1.5+2-3.5=0
\
& t=1: && xibar{S}_1(omega_1)=1.5+3-4.5=0
\
&&& xibar{S}_1(omega_2)=1.5+1-2.5=0
\
&&& xibar{S}_1(omega_3)=1.5+5-5=1.5>0
end{align}$$
Thus:
$$xibar{S}_0=0, quad mathbb{P}(xibar{S}_1geq0)=1, quad mathbb{P}(xibar{S}_1>0)>0$$
Hence the market has arbitrage.
For question b), you need to generalize to prove that there is no portfolio $xi$ that allows arbitrage (instead of just finding a counterexample as in a).
Hello, why is it legit to just take a random $xi$ and show the conditions ? Wouldn't the $xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ?
– Rebellos
Nov 11 at 20:51
It is not a random $xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $xi$, and there is a logic to it.
– Alex C
Nov 11 at 22:48
@Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(omega_1)/S_1^2(omega_1), S_1^1(omega_3)/S_1^2(omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(omega_2)/S_1^1(omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$.
– Daneel Olivaw
Nov 11 at 23:02
Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do?
– Rebellos
Nov 11 at 23:06
add a comment |
The parameter $xi$ represents your strategy, namely the quantity you hold in your portfolio of each security $S^0$, $S^1$ and $S^2$. Consider the following strategy:
$${xi}=(xi^1,xi^2,xi^3)=(1.5,1,-0.5)$$
Then:
$$begin{align}
& t=0: && xibar{S}_0=xi^0S_0^0+xi^1S_0^1+xi^2S_0^2 = 1.5+2-3.5=0
\
& t=1: && xibar{S}_1(omega_1)=1.5+3-4.5=0
\
&&& xibar{S}_1(omega_2)=1.5+1-2.5=0
\
&&& xibar{S}_1(omega_3)=1.5+5-5=1.5>0
end{align}$$
Thus:
$$xibar{S}_0=0, quad mathbb{P}(xibar{S}_1geq0)=1, quad mathbb{P}(xibar{S}_1>0)>0$$
Hence the market has arbitrage.
For question b), you need to generalize to prove that there is no portfolio $xi$ that allows arbitrage (instead of just finding a counterexample as in a).
The parameter $xi$ represents your strategy, namely the quantity you hold in your portfolio of each security $S^0$, $S^1$ and $S^2$. Consider the following strategy:
$${xi}=(xi^1,xi^2,xi^3)=(1.5,1,-0.5)$$
Then:
$$begin{align}
& t=0: && xibar{S}_0=xi^0S_0^0+xi^1S_0^1+xi^2S_0^2 = 1.5+2-3.5=0
\
& t=1: && xibar{S}_1(omega_1)=1.5+3-4.5=0
\
&&& xibar{S}_1(omega_2)=1.5+1-2.5=0
\
&&& xibar{S}_1(omega_3)=1.5+5-5=1.5>0
end{align}$$
Thus:
$$xibar{S}_0=0, quad mathbb{P}(xibar{S}_1geq0)=1, quad mathbb{P}(xibar{S}_1>0)>0$$
Hence the market has arbitrage.
For question b), you need to generalize to prove that there is no portfolio $xi$ that allows arbitrage (instead of just finding a counterexample as in a).
answered Nov 11 at 20:45
Daneel Olivaw
2,8281529
2,8281529
Hello, why is it legit to just take a random $xi$ and show the conditions ? Wouldn't the $xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ?
– Rebellos
Nov 11 at 20:51
It is not a random $xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $xi$, and there is a logic to it.
– Alex C
Nov 11 at 22:48
@Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(omega_1)/S_1^2(omega_1), S_1^1(omega_3)/S_1^2(omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(omega_2)/S_1^1(omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$.
– Daneel Olivaw
Nov 11 at 23:02
Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do?
– Rebellos
Nov 11 at 23:06
add a comment |
Hello, why is it legit to just take a random $xi$ and show the conditions ? Wouldn't the $xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ?
– Rebellos
Nov 11 at 20:51
It is not a random $xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $xi$, and there is a logic to it.
– Alex C
Nov 11 at 22:48
@Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(omega_1)/S_1^2(omega_1), S_1^1(omega_3)/S_1^2(omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(omega_2)/S_1^1(omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$.
– Daneel Olivaw
Nov 11 at 23:02
Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do?
– Rebellos
Nov 11 at 23:06
Hello, why is it legit to just take a random $xi$ and show the conditions ? Wouldn't the $xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ?
– Rebellos
Nov 11 at 20:51
Hello, why is it legit to just take a random $xi$ and show the conditions ? Wouldn't the $xi$ need to be derived from the exercise ? I'm asking because I'm a true beginner at this lesson. Also, for the martingale stuff, what is needed ?
– Rebellos
Nov 11 at 20:51
It is not a random $xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $xi$, and there is a logic to it.
– Alex C
Nov 11 at 22:48
It is not a random $xi$, it is the product of careful thought by someone who knew what he was looking for. You have to construct $xi$, and there is a logic to it.
– Alex C
Nov 11 at 22:48
@Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(omega_1)/S_1^2(omega_1), S_1^1(omega_3)/S_1^2(omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(omega_2)/S_1^1(omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$.
– Daneel Olivaw
Nov 11 at 23:02
@Rebellos I Iooked at the relative prices $S^1/S^2$ and $S^2/S^1$ and noticed that $S_1^1(omega_1)/S_1^2(omega_1), S_1^1(omega_3)/S_1^2(omega_3)>S_0^1/S_0^2$ whereas only $S_1^2(omega_2)/S_1^1(omega_2)>S_0^2/S_0^1$ so I tried to come up with an arbitrage portfolio long the security most expected to increase relatively, namely $S^1$.
– Daneel Olivaw
Nov 11 at 23:02
Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do?
– Rebellos
Nov 11 at 23:06
Thanks for your reply, I get it now! Finally, that martingale stuff, what does it want me to do?
– Rebellos
Nov 11 at 23:06
add a comment |
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