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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?

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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.

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  • $\begingroup$ 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 '15 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 '15 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 '15 at 1:10
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    $\begingroup$ ok got it. I will edit my answer and show that both process are the same. $\endgroup$ – mth_mad Dec 28 '15 at 18:46
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    $\begingroup$ Answer changed above. Hope that helps. $\endgroup$ – mth_mad Dec 28 '15 at 19:14

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