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

$$T &gt; 0$$

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


This is risk-neutral measure.

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

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

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

$\xi_t$ 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,
$$\int_0^T \xi_t dt &lt; \infty \ \text{a.s.} \ \to \ \int_0^T -\xi_t dt &lt; \infty \ \text{a.s.}$$
$$\to L_t := \exp(-\int_0^t (-\xi_s dW_s) - \frac{1}{2} \int_0^t \xi_s^2 ds)$$
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

$$dL_t = L_t \xi_t dW_t$$

$$dM_t = M_t \xi_t dW_t$$


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?
(Looking at the question and notation used more closely, the formulation seems to be problematic in couple places.)

General Fact

Let $W$ be standard Brownian motion with respect to filtration $( \mathscr F_t )_{t \in [0,T]}$. Consider $(L_t)_{t \in [0,T]}$ defined by
$$
\frac{dL_t}{L_t} = \psi_t dL_t, \; L_0 = 1.
$$
In general, $L_t = e^{\int_0^t \psi_s dW_s - \frac12 \int_0^t \psi^2_s ds }$ is a super
martingale. Under some conditions (e.g. Novikov's condition), $L_t$ is a martingale and
one can define a probability measure $\mathbb{Q}$ by
$$
\frac{d \mathbb Q}{d \mathbb P} = L_T.
$$
Under $\mathbb{Q}$, the process
$$
W^{\mathbb{Q}}_t = W_t - \int_0^t \psi_s ds 
$$
is a standard Brownian motion with respect to filtration $( \mathscr F_t )_{t \in [0,T]}$. 

An informal indication why this is true is as follows. Consider $W^{\lambda}_t = W_t + \int_0^t \lambda_s ds$. By Bayes theorem, $W^{\lambda}$ is a $\mathbb{Q}$-martingale if and only if $L W^{\lambda}$ is a $\mathbb{P}$-martingale. Since

\begin{align*}
d L W^{\lambda} &amp;= L d W^{\lambda} + W^{\lambda} dL + dL dW^{\lambda}\\
                &amp;= L (\psi + \lambda) dt + (\cdots) dW,
\end{align*}
we must have $\lambda = - \psi$, for $W^{\lambda}$ to be a $\mathbb{Q}$-Brownian motion.

Discounted Price as Probability Density

The implicit assumptions are that there is a underlying asset whose price $S_t$ follows
$$
\frac{dS_t}{S_t} = r_t dt + \sigma_t dW_t
$$ 
under the risk neutral measure $\mathbb{P}$. The short rate $(r_t)$ and volatility
$\sigma_t$ processes are adapted with sufficient regularity so that the integrals exist. (For this to be true, the Brownian filtration generated by $(W_t)$ under the risk neutral measure has to be the same as that generated by the physical Brownian motion under the physical measure, so that the Martingale Representation Theorem applies.)

In this Brownian filtration setting, for any time-$T$ claim $X_T$, the risk-neutral dynamics of its price $X_t$ takes the form
$$
\frac{d X_t}{X_t} = r_t dt + \psi_t dW_t.
$$
The process $(\psi_t)$ is the volatility of return of $X_t$, under both the physical and risk-neutral measure.

In other words, the risk-neutral dynamics of the discounted price 
$M_t = e^{- \int_0^t r_s ds} X_t$ is given by
$$
\frac{d M_t}{M_t} = \psi_t dW_t, \, M_0 = X_0.
$$
(The discounted price of any $T$-claim must follow a martingale under risk neutral measure, by no arbitrage.)

If Novikov's condition holds, then $L_T = \frac{M_T}{M_0}$ defines a Radon-Nikodym density
$$
\frac{d \mathbb Q}{d \mathbb P} = L_T.
$$
Under $\mathbb{Q}$, the process
$$
W_t - \int_0^t \psi_s ds 
$$
is a standard Brownian motion with respect to filtration $( \mathscr F_t )_{t \in [0,T]}$. 

In other words, the discounted payoff $ e^{- \int_0^T r_s ds} X_T$ of any $T$-claim $X_T$, normalized by its time-$0$ price $X_0$, can be considered as the Radon-Nikodym density of a measure $\mathbb Q$. Under $\mathbb Q$, the risk-neutral Brownian motion now has drift given by the volatility of return $\frac{d X_t}{X_t}$.

If $(Y_t)$ is the price of a traded asset, then 
$ e^{- \int_0^t r_s ds} Y_t$ is a $\mathbb P$-martingale. This implies that $(\frac{Y_t}{X_t})$ is a $\mathbb Q$-martingale.

Forward Measure

The forward measure is a special case of above where $X_t = P(t,T)$ is the time-$t$ price of the zero coupon bond maturing at $T$.
In particular, $X_T = P(T,T) = 1$.
In the expression
$$
\frac{dP(t,T)}{P(t,T)} = r_t dt + \xi_t dW_t,
$$
$\xi_t$ is the volatility of return on zero coupon bond.

(If $(r_t)$ is deterministic, then $\xi = 0$, and the forward measure is the same as the risk neutral measure. The zero-coupon bond is a risky asset only when the short rate is stochastic.)

The corresponding measure $\mathbb Q$ is defined by
$$
\frac{d \mathbb Q}{d \mathbb P} = \frac{e^{- \int_0^T r_s ds} P(T,T)}{P(0,T)} = L_T.
$$
Since
$$
\frac{d L_t}{L_t} = \xi_t dW_t,
$$
it follows from general discussion above that, under $\mathbb{Q}$, the process
$$
W_t - \int_0^t \xi_s ds 
$$
is a standard Brownian motion with respect to filtration $( \mathscr F_t )_{t \in [0,T]}$. 

(In the question posted, the martingale $M_t$ should be $\frac{e^{- \int_0^t r_s ds} P(t,T)}{P(0,T)}$. It's the discounted asset prices that are martingales under risk-neutral measure.)

Empirical Comments 

The forward measure $\mathbb Q$ has the property that the forward prices form a $\mathbb Q$-martingale.

Suppose $F(t,T)$ is the forward price of the forward contract entered at $t$ with maturity $T$. By no-arbitrage (spot-forward parity, in this case) 
$$
F(t,T) P(t,T) = S_t
$$ 
which, after discounting, is a $\mathbb P$-martingale. So $F(t,T)$ is a $\mathbb Q$-martingale.

Since the forward price
$$
F(t,T) = \frac{S_t}{P(t,T)} 
$$
moves inversely relative to the $P(t,T)$.
The forward measure shifts probability mass toward states where 
the discounted return of zero coupon bond
$$
\frac{d \left( e^{- \int_0^t r_s ds} P(t,T) \right)}{ e^{- \int_0^t r_s ds} P(t,T)} = \xi_t dW_t,
$$ 
is high, in such a way that counteracts the movement in $P(t,T)$ and keeps the (conditional) expectation constant.