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 $A\in G$?


1 Answer 1


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 \begin{equation} 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) \end{equation} 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 $S_k = \{\omega |L^{\mathcal{G}}(\omega) = L^{\mathcal{G}}(k)\}$ \begin{equation} \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}) \end{equation} 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: \begin{equation} \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\}) \end{equation} 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 $L^{\mathcal{G}}(\omega \in \{2,3\}) = L^{\mathcal{G}}(2) = L^{\mathcal{G}}(3)$

  • $\begingroup$ I think you are calculating $L$ on $\mathcal F$ while I on $G$. $\endgroup$
    – tosik
    Apr 1, 2019 at 6:22
  • $\begingroup$ I am taking another look. Give me a bit $\endgroup$
    – Slade
    Apr 1, 2019 at 17:08

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