# Introduction of a stochastic discount factor in martingale pricing

The example below is taken from Björk (2009). Let Radon-Nikodym derivative be $$L=\frac{dP}{dQ} \;\; \text{on} \; \mathcal F$$ or written analogously $$P(A) = \int_AL(\omega)dQ(\omega) \;\; \text{for all} \; A\in \mathcal F.$$

For finite sample space $$\Omega$$ this simplifies to $$P(A) = \sum_{\omega\in A}L(\omega)Q(\omega).$$

Let $$\Omega = \{ 1,2,3 \}$$, $$\mathcal F = 2^\Omega$$ and $$G=\left\{ \Omega, \emptyset, \{1\}, \{2,3\} \right\}$$ and $$P(1)=1/4, \;\; P(2)=1/2, \;\; P(3)=1/4,$$ $$Q(1)=1/3, \;\; Q(2)=1/3, \;\; Q(3)=1/3.$$

We are computing $$L$$ on $$G$$. Using above formula for $$A=\{1\}$$ we have that $$1/4 = L(1)\cdot 1/3,$$ which leads to $$L(1)=3/4.$$

Proceeding $$P(\{2,3\}) = L(2)Q(2) + L(3)Q(3),$$ which results in $$L(2) + L(3) = 9/4.$$ Björk writes $$L(2)=9/8$$ and $$L(3)=9/8$$. Could you please explain why $$L(\omega)$$ is constant over the subsets $$A\in G$$?

Let $${\mathcal{F}} = 2^{\Omega}$$ and let $$\mathcal{G} =\left\{ \Omega, \emptyset, \{1\}, \{2,3\} \right\}$$. Both are sigma-algebras of subsets of $$\Omega$$.

The book/question is confusing since the way you have written $$P(A) = \sum_{\omega\in A}L(\omega)Q(\omega)$$ makes it seem like $$Q$$ is a mapping from $$\Omega$$, but probability measures are mappings from sigma-algebras, not the sample space itself. So switching between continuous and discrete makes things more confusing in my opinion, since one is a Lebesgue Integral and the other is a normal summation.

So in the case of $${\mathcal{F}} = 2^{\Omega}$$, since every $$\omega \in \Omega$$ is also a set $$F \in \mathcal{F}$$, then the expression you wrote works fine. So that is why we can write the $$L^{\mathcal{F}}(\omega ) = \frac{P(\omega)}{Q(\omega)}$$ expressions and evaluate them as usual. And we can see that since $$\mathcal{F}$$ is the power set of $$\Omega$$, we know exactly which $$\omega \in \Omega$$ occurs if we know which sets in $$\mathcal{F}$$ occur or do not occur. So $$L^{\mathcal{F}}$$ is measurable with respect to $$\mathcal{F}$$.

In the case of $$L^{\mathcal{G}}$$, the only sets in $$\mathcal{G}$$ are $$\left\{ \Omega, \emptyset, \{1\}, \{2,3\} \right\}$$, so given knowledge of whether those occur we need to be able to calculate $$L^{\mathcal{G}}(\omega)$$. So now assuming that the probability space is equipped with sigma-algebra $$\mathcal{G}$$, we can only calculate $$P(A)$$ & $$Q(A)$$ for $$A \in \left\{ \Omega, \emptyset, \{1\}, \{2,3\} \right\}$$.

So $$$$P(\{2,3\}) = \frac{3}{4} = \int_{\{2,3\}}L^{\mathcal{G}}(\omega)dQ(\omega) = \int_{\Omega}1_{\{2,3\}}(\omega)L^{\mathcal{G}}(\omega)dQ(\omega)$$$$ and imagine that $$L^{\mathcal{G}}(\omega)$$ is a simple function, we can write it as $$L^{\mathcal{G}}(\omega) = L^{\mathcal{G}}(1)1_{\{1\}}(\omega) + L^{\mathcal{G}}(2)1_{\{2\}}(\omega) + L^{\mathcal{G}}(3)1_{\{3\}}(\omega)$$ and so, defining $$S_k = \{\omega |L^{\mathcal{G}}(\omega) = L^{\mathcal{G}}(k)\}$$ $$$$\int_{\Omega}1_{\{2,3\}}(\omega)L^{\mathcal{G}}(\omega)dQ(\omega) = \sum_{k = {1,2,3}}L^{\mathcal{G}}(k)Q(\{2,3\} \cap S_k)\\ = L^{\mathcal{G}}(1)Q(\emptyset) + L^{\mathcal{G}}(2)Q({2}) + L^{\mathcal{G}}(3)Q({3})$$$$ So now we see the issue of $$Q$$ being only defined on $$\left\{ \Omega, \emptyset, \{1\}, \{2,3\} \right\}$$. So the only way for the expression to make sense and thus equal $$P(\{2,3\}) = \frac{3}{4}$$, we need for $$L^{\mathcal{G}}(\omega)$$ as a simple function to be written: $$L^{\mathcal{G}}(\omega) = L^{\mathcal{G}}(1)1_{\{1\}}(\omega) + L^{\mathcal{G}}(2)1_{\{2,3\}}(\omega) = L^{\mathcal{G}}(1)1_{\{1\}}(\omega) + L^{\mathcal{G}}(3)1_{\{2,3\}}(\omega)$$, since then we'd have: $$$$\int_{\Omega}1_{\{2,3\}}(\omega)L^{\mathcal{G}}(\omega)dQ(\omega) = \sum_{k = {1,2}}L^{\mathcal{G}}(k)Q(\{2,3\} \cap S_k)\\ = L^{\mathcal{G}}(1)Q(\emptyset) + L^{\mathcal{G}}(2)Q(\{2,3\})$$$$ which is a valid expression.

So finally $$\frac{3}{4} = L^{\mathcal{G}}(2)Q(\{2,3\}) = L^{\mathcal{G}}(2) \frac{2}{3}$$, so $$L^{\mathcal{G}}(2) = L^{\mathcal{G}}(3) = \frac{9}{8}$$

So the book and this answer is basically a long, convoluted way to say that since $$L^{\mathcal{G}}(\omega)$$ must be $$\mathcal{G}$$-measurable, then the only way to calculate $$L^{\mathcal{G}}(2)$$ and $$L^{\mathcal{G}}(3)$$ with only the knowledge of whether sets in $$\left\{ \Omega, \emptyset, \{1\}, \{2,3\} \right\}$$ occur, is if $$L^{\mathcal{G}}(2) = L^{\mathcal{G}}(3)$$. This is since when $$\{2,3\}$$ occurs, we do not know which of $$2$$ or $$3$$ occurs, just that one of them does. So the only way to know $$L^{\mathcal{G}}(2)$$ and $$L^{\mathcal{G}}(3)$$ is if they are equal. Then we can say that when $$\{2,3\}$$ occurs that $$L^{\mathcal{G}}(\omega \in \{2,3\}) = L^{\mathcal{G}}(2) = L^{\mathcal{G}}(3)$$

• I think you are calculating $L$ on $\mathcal F$ while I on $G$. – tosik Apr 1 at 6:22
• I am taking another look. Give me a bit – Slade Apr 1 at 17:08