Showing that a market model has arbitrage and describing martingales












2














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.










share|improve this question



























    2














    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.










    share|improve this question

























      2












      2








      2







      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.










      share|improve this question













      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






      share|improve this question













      share|improve this question











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      share|improve this question










      asked Nov 11 at 18:10









      Rebellos

      1284




      1284






















          1 Answer
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          oldest

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          2














          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).






          share|improve this answer





















          • 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











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          1 Answer
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          1 Answer
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          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2














          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).






          share|improve this answer





















          • 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
















          2














          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).






          share|improve this answer





















          • 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














          2












          2








          2






          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).






          share|improve this answer












          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).







          share|improve this answer












          share|improve this answer



          share|improve this answer










          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


















          • 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


















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