Jtdylktuy

Exercise :

Consider the finite sample space $Omega = {omega_1, omega_2, omega_3}$ and the probability space $(Omega, mathcal{F}, mathbb P)$ with $mathcal F:= 2^Omega$ and let $mathbb P$ be a probability measure such that $mathbb P[{omega_1}] > 0$ for all $i=1,2,3$ .

Let :
$$bar{S}_0 = begin{pmatrix} 1\2\7 end{pmatrix}, quadbar{S}_1(omega_1) = begin{pmatrix} 1\3\9end{pmatrix}, quad bar{S}_1(omega_2) = begin{pmatrix} 1\1\5end{pmatrix}, quad bar{S}_1(omega_3) = begin{pmatrix} 1\13\10 end{pmatrix}$$
Describe all the equivalent martingale measures.

Definitions and request for help/elaboration :

Theorem : Let $mathbb P, mathbb Q$ be two probability measures over the space $(Omega, mathcal{F})$. Then, it is $mathbb P sim mathbb Q$ if and only if there exists a random variable $X>0, ; X in L^1(mathbb P)$ such that :
$$mathbb E_mathbb Q[mathbf{1}_A] = mathbb Q(A) = int_AXmathrm{d}mathbb P=mathbb E_mathbb P[Xmathbf{1}_A], quad forall A in mathcal F$$

Definition : A probability measure $mathbb Q$ over the space $(Omega, mathcal{F})$ is called a martingale measure (MM) if it is :
$$S_1 in L^1(mathbb Q) quad text{and} quad S_0 = mathbb{E}_mathbb Qbigg[frac{S_1}{1+r}bigg]$$

Definition : If $mathbb Q$ is a martingale measure and it also is $mathbb Q sim mathbb P$, then $mathbb Q$ is called an equivalent martingale measure (EMM).

Definition : We define $mathcal{P}$ to be the set of all the equivalent martingale measures :
$$mathcal{P} = {mathbb Q|mathbb Q sim mathbb P, mathbb Q ; text{is an equivalent martingale measure}}$$

Question : Now, given these definitions, what is needed for me to carry out so I can answer properly to the exercise given ? What am I expected to do and how ? I would really appreciate a thorough elaboration as I haven't been properly introduced to martingales and measure theory yet (there are just tools needed for a higher-level class).

From the fact that the first asset does not change value, we can infer that the interest rate is zero. The problem is to find all probability measures $Q$ such that $$ S_0 = S_1Q(omega_1) + S_2Q(omega_2) +S_3Q(omega_3),$$ and all of the three probabilities are greater than zero (and add up to one, of course). The equation expresses the fact that it is a martingale measure (the RHS is the expected discounted value). It needs all probabilities greater than zero in order to be equivalent to $P.$

As far as I can tell, there aren’t any (I find one probability measure that has the right expected value but it doesn’t have all probabilities nonzero.)

Edit

Going to expand on equivalence a bit. Two measures are equivalent if they have the same null events. I notice the part on equivalence is marked as a theorem, not a definition, so I suspect perhaps you were given this definition at at some point. Since the physical measure $P$ you've been given has $P(omega)>0$ for all $omegainOmega,$ the only null event is the empty set. Thus for $Q$ to be equivalent, it must also have the only null event be the emptyset, i.e. $Q(omega)>0$ for all $omegain Omega.$

To see this is equivalent to the theorem that there exists a random variable $X>0$ such that for any event $A$ $$ E_Q(1_A) = E_P(X1_A),$$ as I mentioned in the comments, just take $X(omega) = frac{Q(omega)}{P(omega)},$ for all $omega$ such that $P(omega) >0,$ and let it be defined as any positive number elsewhere. We see that the fact that P and Q are equivalent implies $X>0$. And we have (specializing to the sample space in the problem for clarity) $$ E_Q(1_A) = 1_A(omega_1)Q(omega_1) + 1_A(omega_2)Q(omega_2) + 1_A(omega_3)Q(omega_3) \=1_A(omega_1)X(omega_1)P(omega_1) + 1_A(omega_2)X(omega_2)P(omega_2) + 1_A(omega_3)X(omega_3)P(omega_3)\=E_P(1_A X) $$ where we used equivalence again to deal with the possibility that $P(omega_i) = 0$ for some $i$.

In the reverse direction, for any $X>0$, if there is an $A$ such that $Q(A) =0$ but $P(A)ne 0,$ or vice versa then it's clear that $E_Q(1_A) ne E_P(X 1_A)$ since one side is zero and the other isn't $(E_Q(1_A) = Q(A)$ and $E_P(X1_A) = 0$ if and only if $P(A) = 0$ since $X>0.$)