Let $T > 0$, and let $(\Omega, \mathscr F, \{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)$ be a filtered probability space where $\mathbb P = \tilde{\mathbb P}$ (risk-neutral measure) and $\mathscr F_t = \mathscr F_t^{{W}} = \mathscr F_t^{\tilde{W}}$ where $W = \tilde{W} = (\tilde{W_t})_{t \in [0,T]} = ({W_t})_{t \in [0,T]}$ is standard $\mathbb P=\tilde{\mathbb P}$-Brownian motion.

Define forward measure $\hat{\mathbb P}$:

$$A_T := \frac{d \hat{\mathbb P}}{d \mathbb P} = \frac{\exp(-\int_0^T r_s ds)}{P(0,T)}$$

It can be shown that $\exp(-\int_0^t r_s ds)P(t,T)$ is a $(\mathscr F_t, \mathbb P)-$martingale where $r_t$ is short rate process and $P(t,T)$ is bond price.

We are given that

$$\frac{dP(t,T)}{P(t,T)} = r_t dt + \zeta_t dW_t$$

where $r_t$ and $\zeta_t$ are $\mathscr F_t$-adapted and $\zeta_t$ satisfies Novikov's condition. I don't think $\zeta_t$ is supposed to represent anything in particular.

Define the stochastic process $\hat{W} = (\hat{W_t})_{t\in[0,T]}$ s.t.

$$\hat{W_t} := W_t + \int_0^t -\zeta_s ds$$

Use Girsanov Theorem to prove $\hat{W_t}$ is standard $\hat{\mathbb P}$-Brownian motion.

What I tried:

Since $\zeta_t$ satisfies Novikov's condition, $\int_0^T -\zeta_t dt < \infty$ a.s. and

$$L_t := \exp(-\int_0^t (-\zeta_s dW_s) - \frac{1}{2} \int_0^t \zeta_s^2 ds)$$

is a $(\mathscr F_t, \mathbb P)-$martingale.

By Girsanov Theorem, $\hat{W_t}$ is standard $\mathbb P^{*}$-Brownian Motion where

$$\frac{d \mathbb P^{*}}{d \mathbb P} = L_T$$

I guess we have that $\hat{W_t}$ is standard $\hat{\mathbb P}$-Brownian Motion if we can show that

$$L_T = \frac{d \hat{\mathbb P}}{d \mathbb P}$$

I think I was able to show (lost my notes) that $dL_t = L_t \zeta_t dW_t$, $dA_t = A_t \zeta_t dW_t$ and then $d(\ln L_t) = d(\ln A_t)$

From $d(\ln L_t) = d(\ln A_t)$, I infer that $L_t = A_t$ and hence $L_T = A_T$ QED.

Is that right?


1 Answer 1


Your notations are really hard to follow as you define $\mathbb{P}$ twice at the beginning. The notation $\mathbb{P} = \mathbb{\hat{P}}$ and $\mathbb{P} =\mathbb{\tilde{P}}$ is not meaningful as the probability measure $\mathbb{P}$ is already fixed and used for the real world probability measure. I think that this is the reason why you are getting confused.

Here is the solution. I am using standard notations here. Under $\mathbb{Q}$ the risk neutral probability $$\frac{d P_{tT}}{P_{tT}} = r_t dt + \xi_t dW_t$$

Now consider the process $\displaystyle Z_t = \exp(-\int_{0}^t r_s ds)\frac{P_{tT}}{P_{0T}}$. Note that with your notation $Z_T = A_T$, since $P_{TT} = 1$.

If we show that is a $Z_t$ is a $\mathbb{Q}$-martingale, with $Z_0 = 1$, then we can apply a change of measure to define $\mathbb{Q}_{T}$, the forward measure, as $$\mathbb{Q}_{T}(B) = \mathbb{E}_{\mathbb{Q}}(Z_T \cdot I_{B}),$$ for $B \in \mathcal{F}_T$. Then, by Girsanov theorem, $\hat{W}_t = W_t -\int_0^t \xi_s ds$ is a B.M under $\mathbb{Q}_T$. Note the minus sign and not the plus sign as in question.

Proof that $Z_t$ is a martingale with $Z_0 = 1$: The fact that $Z_0 = 1$ is clear. For the martingale property, we have that from the dynamics of $P_{tT}$ under $\mathbb{Q}$, \begin{align*} d(\exp(-\int_{0}^t r_s ds)P_{tT}) = \exp(-\int_{0}^t r_s ds)P_{tT} \xi_t dW_t \end{align*} Hence, $dZ_t = Z_t \xi_t dW_t$ or equivalently by taking the log and apply Ito's formula, $$Z_t = \exp\left( \int_{0}^{t} \xi_s dW_s - \frac{1}{2} \int_{0}^{t} \xi^2_s ds\right)$$ Note that here $Z_t = L_t$. As we are told that it verifies Novikov condition, this ensures that it is a martingale and that we can apply Girsanov.

  • $\begingroup$ Thanks mth_mad. Edited. To what does your $Z_t$ correspond in my solution? $A_t$? $L_t$? I think somewhere in your solution you assumed something I was trying to prove (ie $A_t = L_t$) $\endgroup$
    – BCLC
    Dec 27, 2015 at 9:13
  • $\begingroup$ Yes my $Z_t$ is the $A_t$ given in the question, which you have rewritten $L_t$ in your solution. I don't see why you are using different notation as they are equal by construction. Finally, the "seems to imply" statement in your solution is Girsanov theorem. No magic trick here ;-) $\endgroup$
    – mth_mad
    Dec 27, 2015 at 23:55
  • $\begingroup$ right thanks edited. What do you mean by equal by construction? Um, is my proof correct or not? Actually in your proof you claimed Q is forward measure. How do you know that? I think that is precisely what I am trying to prove $\endgroup$
    – BCLC
    Dec 28, 2015 at 1:10
  • 1
    $\begingroup$ ok got it. I will edit my answer and show that both process are the same. $\endgroup$
    – mth_mad
    Dec 28, 2015 at 18:46
  • 1
    $\begingroup$ Answer changed above. Hope that helps. $\endgroup$
    – mth_mad
    Dec 28, 2015 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.