# Show that $W_t - \int_0^t \xi_s ds$ is forward-measure-Brownian

Definitions and stuff:

Consider a filtered probability space $(\Omega, \mathscr F, \{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)$ where

1. $$T > 0$$

2. $$\mathbb P = \tilde{\mathbb P}$$

This is risk-neutral measure.

1. $$\mathscr F_t = \mathscr F_t^{{W}} = \mathscr F_t^{\tilde{W}}$$

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.

Consider $M = \{M_t\}_{t \in [0,T]}$ where

$$M_t := \frac{\exp(-\int_0^t r_s ds)}{P(0,t)}$$

Define forward measure $\mathbb Q$:

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

where $\{r_t\}_{t \in [0,T]}$ is short rate process and $\{P(t,T)\}_{t \in [0,T]}$ is bond price at time t.

It can be shown that $\{\exp(-\int_0^t r_s ds)P(t,T)\}_{t \in [0,T]}$ is a $(\mathscr F_t, \mathbb P)-$martingale where bond price dynamics are given as:

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

where

1. $r_t$ and $\xi_t$ are $\mathscr F_t$-adapted

2. $\xi_t$ satisfies Novikov's condition (I don't think $\xi_t$ is supposed to represent anything in particular)

Problem:

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

$$W_t^{\mathbb Q} := W_t - \int_0^t \xi_s ds$$

Use Girsanov Theorem to prove:

$$W_t^{\mathbb Q} \ \text{is standard} \ \mathbb Q \ \text{-Brownian motion.}$$

What I tried:

Since $\xi_t$ satisfies Novikov's condition,

$$\int_0^T \xi_t dt < \infty \ \text{a.s.} \ \to \ \int_0^T -\xi_t dt < \infty \ \text{a.s.}$$

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

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

By Girsanov Theorem,

$$W_t^{\mathbb Q} \ \text{is standard} \ \mathbb P^{*} \ \text{-Brownian motion, where}$$

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

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

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

I lost my notes, but I think I was able to show using Ito's lemma that

1. $$dL_t = L_t \xi_t dW_t$$

2. $$dM_t = M_t \xi_t dW_t$$

From those I infer that

$$d(\ln L_t) = d(\ln M_t)$$

$$\to L_t = M_t$$

$$\to L_T = M_T$$

QED

Is that right?

• Why is the bond price discounted by the short rate a P-martingale? Your bond price is a generalized GBM. Write it as the exponential of an Ito diffusion, one should see that discounting by the short rate does not account for the Ito correction. – Michael Apr 13 '17 at 16:41
• @Michael are you sure you mean P as in risk neutral and not P as in real world? – BCLC Apr 13 '17 at 16:48
• Ok, I see. If you solve the SDE for $P_t$ as an Ito exponential then substitute into $M_T$, you'll see that Girsanov's theorem applies immediately. Also, $\frac{dL}{L}$ and $d \ln L$ are not the same in the Ito setting. In your argument, one should invoke the uniqueness of strong solutions of SDE's instead. – Michael Apr 14 '17 at 15:09
• @Michael Thanks! Which part of the argument exactly? – BCLC Mar 23 '18 at 8:57