# How can I use the Radon-Nikodym theorem to show that forward measure is indeed measure?

The following statements are taken from the Wikipedia page for forward measure.

Let $$B(T)=\exp \left(\int _{0}^{T}r(u)\,du\right)$$ be the bank account or money market account numeraire and $$D(T)=1/B(T)=\exp \left(-\int _{0}^{T}r(u)\,du\right)$$ be the discount factor in the market at time 0 for maturity T. If $$Q_{*}$$ is the risk neutral measure, then the forward measure $$Q_{T}$$ is defined via the Radon–Nikodym derivative given by $$\frac{dQ_{T}}{dQ_{*}}={\frac {1}{B(T)E_{Q_{*}}[1/B(T)]}}={\frac {D(T)}{E_{Q_{*}}[D(T)]}}.$$

How can I use the Radon-Nikodym theorem to prove that $$Q_T$$ defined above is indeed a measure?

• What is your question exactly: to verify that $Q_T$ has the properties of of a probability measure (i.e. function image in $[0,1]$, and countable additivity of disjoint sets)? – Daneel Olivaw Apr 6 at 13:07
• @DaneelOlivaw Yes. – Idonknow Apr 6 at 13:08

Just to briefly add to Daneel's top answer, start with $$Q_T[A]:=E_{Q_*}\left[1_A \frac{D(T)}{E_{Q_*}[D(T)]}\right].$$

1. Since $$D(T)>0$$, we have that $$Q_T$$ is always non-negative. As usual, $$Q_T[\emptyset]=0$$ and $$Q_T[\Omega]=1$$.

2. Let $$A_1,A_2,...$$ be a sequence of disjoint sets taken from $$\mathcal{F}$$. Then, \begin{align*} Q_T\left[\bigcup_{k=1}^\infty A_k\right] &= E_{Q_*}\left[1_{\bigcup_{k=1}^\infty A_k}\frac{D(T)}{E_{Q_*}[D(T)]}\right] \\ &= \sum_{k=1}^\infty E_{Q_*}\left[1_{A_k}\frac{D(T)}{E_{Q_*}[D(T)]}\right] \\ &= \sum_{k=1}^\infty Q_T\left[1_{A_k}\right], \end{align*} where the second equality stems from splitting up the integral domain.

Introduction

Technically, I don't think you need the Radon-Nikodym theorem here. That theorem assumes the existence of two equivalent probability measures $$Q_1$$ and $$Q_2$$ and states that there must exist a random variable $$\xi$$ such that $$Q_2$$ is defined as the expectation of $$\xi$$ under $$Q_1$$. What you need here is more akin to Theorem 1.6.1 in Shreve (2004), namely given a measure $$Q_1$$ and a random variables $$\xi$$, prove that you can construct a well-defined probability measure $$Q_2$$.

Let $$(\Omega,\mathcal{F},Q_*)$$ be a probability space equipped with the filtration $$\{\mathcal{F}_t\}_{t\geq 0}$$, where $$Q_*$$ is the risk-neutral measure. $$B(t)$$ is defined as the money market account, and $$P(t,T)$$ as the zero-coupon bond with maturity $$0\leq t\leq T$$. We have: $$P(0,T)=E^{Q_*}\left(\left.\frac{B(0)}{B(T)}\right.\right)$$ By definition, $$B(t)>0$$, which implies $$P(t,T)>0$$. Let us define the random-variable $$\xi$$: $$\xi:=\frac{B(0)P(T,T)}{B(T)P(0,T)}$$ By the preceding, the random variable $$\xi$$ is strictly positive. Additionally, under the risk-neutral measure $$Q_*$$, $$\xi$$ has expectation $$1$$ by the martingale property of discounted payoffs: $$E^{Q_*}\left(\xi\right)=\frac{B(0)}{P(0,T)}E^{Q_*}\left(\frac{P(T,T)}{B(T)}\right)=1$$ Hence $$\xi$$ is a valid Radon-Nikodym derivative and we can define the $$T$$-forward measure $$Q_T$$ as follows, for any $$F\in\mathcal{F}$$: $$Q_T(F):=E^{Q_*}\left(\xi 1_{F}\right)=\int_{\omega\in F}\xi(\omega) dQ_*(\omega)$$

1) Image in $$[0,1]$$: note that, for any $$F\in\mathcal{F}$$: $$0 \leq 1_F \leq 1_\Omega$$ Thus: $$0\leq E^{Q_*}\left(\xi 1_F\right)\leq E^{Q_*}\left(\xi 1_\Omega\right)=\int_{\omega\in \Omega}\xi(\omega) dQ_*(\omega)=E^{Q_*}\left(\xi\right)=1$$

2) Countable additivity of disjoint sets: note that, for any $$F_1,F_2\in\mathcal{F}$$ such that $$F_1\cap F_2=\emptyset$$: $$1_{F_1\cup F_2}=1_{F_1}+1_{F_2}-1_{F_1\cap F_2} = 1_{F_1}+1_{F_2}$$ which generalizes. Thus for an infinite, countable sequence of events $$F_1, F_2, \dots$$, you can use the fact that $$0\leq 1_{\cup_{n>0} F_n}<2$$ to invoke the dominated convergence theorem and conclude that: \begin{align}\sum_{n>0}Q_T(F_n) =\lim_{n\rightarrow\infty}\sum_{i\leq n}\int_\Omega\xi(\omega)1_{F_i}(\omega)dQ_*(\omega) &=\lim_{n\rightarrow\infty}\int_\Omega\xi(\omega)1_{\cup_{i\leq n}F_i}(\omega)dQ_*(\omega) \\ &=\int_\Omega\xi(\omega)1_{\cup_{n>0} F_n}(\omega)dQ_*(\omega) \\[8pt] &=Q_T(\cup_{n>0} F_n) \end{align}

You can extend the Radon-Nikodym derivative to any time $$t\in(0,T]$$ by constructing the Radon-Nikodym derivative process. This is done via the conditional expectation: $$\xi(t):=E^{Q_*}\left(\xi|\mathcal{F}_t\right)=E^{Q_*}\left(\left.\frac{B(0)P(T,T)}{B(T)P(0,T)}\right|\mathcal{F}_t\right)=\frac{B(0)P(t,T)}{B(t)P(0,T)},$$ where we have used the fact that any traded-asset rebased by the money market account is a martingale under $$Q_*$$. You can easily verify the properties proved for $$t=0$$ are carried over.

References

Steven Shreve. Stochastic Calculus in Finance II: Continuous Time Models. Springer, 2004.

• Maybe I am missing something, but I feel like this way overcomplicates the answer. If you have a measure $\mu$ on some measurable space $\Omega$ and a non-negative measurable function $f,$ then $\nu(A) = \int_A f$ will define always measure. In this case, it's also obvious, that $\int_\Omega f = 1,$ so it's a probability measure. – LazyCat Apr 6 at 15:18
• @LazyCat what you say is correct, but it is not straightforward: it is actually a theorem. What I am doing is displaying some steps on how the underlying argument goes. I am also showing why the ratio of numéraires is a well-defined Radon-Nikodym derivative. I am also making clear the construction of the RN derivative along t. – Daneel Olivaw Apr 6 at 15:29
• In particular, the OP asked about how to verify $Q_T$ is a well-defined probability measure using the RN theorem, but the RN theorem is unapplicable in that context. – Daneel Olivaw Apr 6 at 15:34
• Well, my point is that unless I am mistaken your response is much more complicated than needs be. For your second question - this is precisely how the denominator is in OP's formula is chosen - to normalize the whole thing to 1. – LazyCat Apr 6 at 16:31
• I certainly didn't mean it to become a pissing contest. I was surprised by your rather involved answer to what seemed to be such a simple question, and assumed I am missing something obvious. Doesn't seem to be the case so far. Yes, you are reading misreading my comments. In the first comment, I meant the function $f$ in the OP's question which is already normalized, in the later one I meant a general function, which still needs to be normalized. – LazyCat Apr 6 at 19:13

If you have an integrable measurable non-negative $$f$$ on a probability space $$(\Omega, \mu),$$ the $$\nu(A) = \int_A f d\mu$$ always defines a measure on $$\Omega.$$ In particular, there's no need to check things like countable additivity, it simply follows from the corresponding properties of the integral.
In general, if $$\int_\Omega f d\mu > 0,$$ one can normalize this measure be a probability measure: $$\nu(A) = \frac1Z \int_A f d\mu,$$ where $$Z = \int_\Omega f d\mu$$
In this case $$f$$ referred as a Radon-Nikodym derivative is already normalized, so you get a probability measure.