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?

  • $\begingroup$ 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)? $\endgroup$ – Daneel Olivaw Apr 6 '20 at 13:07
  • $\begingroup$ @DaneelOlivaw Yes. $\endgroup$ – Idonknow Apr 6 '20 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.



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 $Q_2$.

Radon-Nikodym Derivative

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: $$\begin{align}\sum_{n>0}Q_T(F_n) =\lim_{n\rightarrow\infty}\sum_{i\leq n}\int_\Omega\xi(\omega)1_{F_i}(\omega)dQ_*(\omega) &=\lim_{n\rightarrow\infty}\int_\Omega\xi(\omega)1_{\cup_{i\leq n}F_i}(\omega)dQ_*(\omega) \\ &=\int_\Omega\xi(\omega)1_{\cup_{n>0} F_n}(\omega)dQ_*(\omega) \\[8pt] &=Q_T(\cup_{n>0} F_n) \end{align}$$

Radon-Nikodym Derivative Process

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.


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

  • 2
    $\begingroup$ 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. $\endgroup$ – LazyCat Apr 6 '20 at 15:18
  • $\begingroup$ @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. $\endgroup$ – Daneel Olivaw Apr 6 '20 at 15:29
  • $\begingroup$ 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. $\endgroup$ – Daneel Olivaw Apr 6 '20 at 15:34
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    $\begingroup$ 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. $\endgroup$ – LazyCat Apr 6 '20 at 16:31
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    $\begingroup$ 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. $\endgroup$ – LazyCat Apr 6 '20 at 19:13

Since my comment thread to Daneel's answer became even longer than his original answer, I thought of putting my comments here separately.

I believe, that wiki quotation refers is the following simple fact:

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.


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