The Radon-Nikodym derivative $\frac{dQ_T}{dQ_B}$ to change the measure from $Q_B$ to $Q_T$ can be obtained by considering the expectations

$\mathbb{E}^{Q_T} [\frac{P(t; T)V(T)}{P(T; T)}|\mathcal{F}_t]= \mathbb{E}^{Q_B} [\frac{P (t; T)V (T)}{P(T; T)}\frac{dQ_T}{dQ_B} |\mathcal{F}_t]=\mathbb{E}^{Q_B}[\frac{B(t)V(T)}{B(T)}|\mathcal{F}_t]$

where the first equality comes from applying the Radon-Nikodym derivative so that $E^{Q_T} [X] = E^{Q_B}[X \frac{dQ_T}{dQ_B}]$, and the second equality comes from Equation (11).

Hence we have $\frac{dQ_T}{dQ_B} =\frac{B(t)/B(T)}{P(t; T)/P(T; T)}$.

I do not understand why the bold part hold. It is concluded from $\mathbb{E}^{Q_B} [\frac{P (t; T)V (T)}{P(T; T)}\frac{dQ_T}{dQ_B} |\mathcal{F}_t]=\mathbb{E}^{Q_B}[\frac{B(t)V(T)}{B(T)}|\mathcal{F}_t]$ that $\frac{dQ_T}{dQ_B} =\frac{B(t)/B(T)}{P(t; T)/P(T; T)}$.

Here $(\mathcal{F}_t)_{t\geq 0}$ is a filtration.

We have that two $\mathcal{F}_t$ random variables which have the same conditional expectation for every $t<T$. Why does this imply that the two random variables are equal (a.s.)?

  • 1
    $\begingroup$ This is almost the definition of a question "missing context or other details". What is $P(\cdot,\cdot)$? What is $V$? What is $\mathcal F_t$? What is equation (11)? $\endgroup$
    – Jason
    Commented Nov 29, 2017 at 18:11
  • $\begingroup$ @Jason Equation (11) is not really relevant... My question is basically why can it be deduced from $E[X|\mathcal{F}_t]=E[Y|\mathcal{F}_t]$ for $t\leq T$ that $X=Y$ $\endgroup$
    – Lost1
    Commented Nov 30, 2017 at 9:50

1 Answer 1


We assume that $\frac{dQ_T}{dQ_B}$ is $\mathcal{F}_T$ measurable. Note that, by assumption, for any $V_T \in \mathcal{F}_T$, \begin{align*} \mathbb{E}^{Q_B} \left(\frac{P (t; T)V (T)}{P(T; T)}\frac{dQ_T}{dQ_B} \,\big|\,\mathcal{F}_t\right)=\mathbb{E}^{Q_B}\left(\frac{B(t)V(T)}{B(T)}\,\big|\,\mathcal{F}_t\right). \end{align*} Let \begin{align*} V_T = \frac{P (t; T)}{P(T; T)} \frac{dQ_T}{dQ_B} - \frac{B(t)}{B(T)}. \end{align*} Then, \begin{align*} \mathbb{E}^{Q_B} \left( \left[\frac{P (t; T)}{P(T; T)} \frac{dQ_T}{dQ_B} - \frac{B(t)}{B(T)}\right]^2\,\big|\,\mathcal{F}_t\right) = 0. \end{align*} That is, \begin{align*} \frac{P (t; T)}{P(T; T)} \frac{dQ_T}{dQ_B} - \frac{B(t)}{B(T)} =0. \end{align*} Consequently, \begin{align*} \frac{dQ_T}{dQ_B} = \frac{B(t)P(T; T)}{B(T)P (t; T)}. \end{align*}


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