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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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