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

<p><strong>General Fact</strong></p>

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

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

<p><span class="math-container">\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*}</span>
we must have <span class="math-container">$\lambda = - \psi$</span>, for <span class="math-container">$W^{\lambda}$</span> to be a <span class="math-container">$\mathbb{Q}$</span>-Brownian motion.</p>

<p><strong>Discounted Price as Probability Density</strong></p>

<p>The implicit assumptions are that there is a underlying asset whose price <span class="math-container">$S_t$</span> follows
<span class="math-container">$$
\frac{dS_t}{S_t} = r_t dt + \sigma_t dW_t
$$</span> 
under the risk neutral measure <span class="math-container">$\mathbb{P}$</span>. The short rate <span class="math-container">$(r_t)$</span> and volatility
<span class="math-container">$\sigma_t$</span> processes are adapted with sufficient regularity so that the integrals exist. (For this to be true, the Brownian filtration generated by <span class="math-container">$(W_t)$</span> 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.)</p>

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

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

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

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

<p>If <span class="math-container">$(Y_t)$</span> is the price of a traded asset, then 
<span class="math-container">$ e^{- \int_0^t r_s ds} Y_t$</span> is a <span class="math-container">$\mathbb P$</span>-martingale. This implies that <span class="math-container">$(\frac{Y_t}{X_t})$</span> is a <span class="math-container">$\mathbb Q$</span>-martingale.</p>

<p><strong>Forward Measure</strong></p>

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

<p>(If <span class="math-container">$(r_t)$</span> is deterministic, then <span class="math-container">$\xi = 0$</span>, 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.)</p>

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

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

<p><strong>Empirical Comments</strong> </p>

<p>The forward measure <span class="math-container">$\mathbb Q$</span> has the property that the forward prices form a <span class="math-container">$\mathbb Q$</span>-martingale.</p>

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

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

CC BY-SA 4.0
2020-03-26T02:18:08.273Z
Well, there are some gaps in your argument. 1. Novikov's condition is not quoted correctly. 2. The intended RN density process $M_t$ is not defined correctly. 3. After Ito's lemma is used, taking logs is fine but the result already follows from uniqueness of solutions to SDE.
Michael
CC BY-SA 4.0
2020-03-17T03:14:06.827Z
2020-03-18T01:32:08.930Z
2020-03-18T01:32:08.930Z
Michael
<p><strong>Definitions and stuff:</strong></p>
<p>Consider a filtered probability space <span class="math-container">$(\Omega, \mathscr F, \{\mathscr F_t\}_{t \in [0,T]}, \mathbb P)$</span> where</p>
<ol>
<li><p><span class="math-container">$$T &gt; 0$$</span></p>
</li>
<li><p><span class="math-container">$$\mathbb P = \tilde{\mathbb P}$$</span></p>
</li>
</ol>
<p>This is <a href="https://en.wikipedia.org/wiki/Risk-neutral_measure" rel="nofollow noreferrer">risk-neutral measure</a>.</p>
<ol start="3">
<li><span class="math-container">$$\mathscr F_t = \mathscr F_t^{{W}} = \mathscr F_t^{\tilde{W}}$$</span></li>
</ol>
<p>where <span class="math-container">$W = \tilde{W} = \{\tilde{W_t}\}_{t \in [0,T]} = \{{W_t}\}_{t \in [0,T]}$</span> is standard <span class="math-container">$\mathbb P=\tilde{\mathbb P}$</span>-Brownian motion.</p>
<p>Consider <span class="math-container">$M = \{M_t\}_{t \in [0,T]}$</span> where</p>
<p><span class="math-container">$$M_t := \frac{\exp(-\int_0^t r_s ds)}{P(0,t)}$$</span></p>
<p>Define <a href="https://en.wikipedia.org/wiki/Forward_measure" rel="nofollow noreferrer">forward measure</a> <span class="math-container">$\mathbb Q$</span>:</p>
<p><span class="math-container">$$\frac{d \mathbb Q}{d \mathbb P} := M_T = \frac{\exp(-\int_0^T r_s ds)}{P(0,T)}$$</span></p>
<p>where <span class="math-container">$\{r_t\}_{t \in [0,T]}$</span> is short rate process and <span class="math-container">$\{P(t,T)\}_{t \in [0,T]}$</span> is bond price at time t.</p>
<p>It can be shown that <span class="math-container">$\{\exp(-\int_0^t r_s ds)P(t,T)\}_{t \in [0,T]}$</span> is a <span class="math-container">$(\mathscr F_t, \mathbb P)-$</span>martingale where bond price dynamics are given as:</p>
<p><span class="math-container">$$\frac{dP(t,T)}{P(t,T)} = r_t dt + \xi_t dW_t$$</span></p>
<p>where</p>
<ol>
<li><p><span class="math-container">$r_t$</span> and <span class="math-container">$\xi_t$</span> are <span class="math-container">$\mathscr F_t$</span>-adapted</p>
</li>
<li><p><span class="math-container">$\xi_t$</span> satisfies Novikov's condition (I don't think <span class="math-container">$\xi_t$</span> is supposed to represent anything in particular)</p>
</li>
</ol>
<hr />
<p><strong>Problem:</strong></p>
<blockquote>
<p>Define the stochastic process <span class="math-container">$W^{\mathbb Q} = (W_t^{\mathbb Q})_{t\in[0,T]}$</span> s.t.</p>
<p><span class="math-container">$$W_t^{\mathbb Q} := W_t - \int_0^t \xi_s ds$$</span></p>
<p>Use <strong><a href="https://en.wikipedia.org/wiki/Girsanov_theorem" rel="nofollow noreferrer">Girsanov Theorem</a></strong> to prove:</p>
<p><span class="math-container">$$W_t^{\mathbb Q} \ \text{is standard} \ \mathbb Q \ \text{-Brownian motion.}$$</span></p>
</blockquote>
<hr />
<p><strong>What I tried:</strong></p>
<p>Since <span class="math-container">$\xi_t$</span> satisfies Novikov's condition,</p>
<p><span class="math-container">$$\int_0^T \xi_t dt &lt; \infty \ \text{a.s.} \ \to \ \int_0^T -\xi_t dt &lt; \infty \ \text{a.s.}$$</span></p>
<p><span class="math-container">$$\to L_t := \exp(-\int_0^t (-\xi_s dW_s) - \frac{1}{2} \int_0^t \xi_s^2 ds)$$</span></p>
<p>is a <span class="math-container">$(\mathscr F_t, \mathbb P)-$</span>martingale.</p>
<p>By Girsanov Theorem,</p>
<p><span class="math-container">$$W_t^{\mathbb Q} \ \text{is standard} \ \mathbb P^{*} \ \text{-Brownian motion, where}$$</span></p>
<p><span class="math-container">$$\frac{d \mathbb P^{*}}{d \mathbb P} := L_T$$</span></p>
<p>I guess we have that <span class="math-container">$W_t^{\mathbb Q}$</span> is standard <span class="math-container">$\mathbb Q$</span>-Brownian Motion if we can show that</p>
<p><span class="math-container">$$L_T = \frac{d \mathbb Q}{d \mathbb P}$$</span></p>
<p>I lost my notes, but I think I was able to show using Ito's lemma that</p>
<ol>
<li><p><span class="math-container">$$dL_t = L_t \xi_t dW_t$$</span></p>
</li>
<li><p><span class="math-container">$$dM_t = M_t \xi_t dW_t$$</span></p>
</li>
</ol>
<p>From those I infer that</p>
<p><span class="math-container">$$d(\ln L_t) = d(\ln M_t)$$</span></p>
<p><span class="math-container">$$\to L_t = M_t$$</span></p>
<p><span class="math-container">$$\to L_T = M_T$$</span></p>
<p>QED</p>
<p>Is that right?</p>

CC BY-SA 3.0
2017-04-13T16:41:49.460Z
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
CC BY-SA 3.0
2017-04-13T16:48:29.193Z
@Michael are you sure you mean P as in risk neutral and not P as in real world?
BCLC
CC BY-SA 3.0
2017-04-14T15:09:49.213Z
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
CC BY-SA 3.0
2018-03-23T08:57:36.050Z
@Michael Thanks! Which part of the argument exactly?
BCLC
<mathematical-economics><finance><financial-economics><probability><bonds>
Show that $W_t - \int_0^t \xi_s ds$ is forward-measure-Brownian
BCLC
# Show that $W_t - \int_0^t \xi_s ds$ is forward-measure-Brownian

(Asked by: username_0 on 2015-12-28)

**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](https://en.wikipedia.org/wiki/Risk-neutral_measure).

3. $$\mathscr F\_t = \mathscr F\_t^{{W}} = \mathscr F\_t^{\tilde{W}}$$

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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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?

**Tags:** mathematical-economics, finance, financial-economics, probability, bonds
> **Comments:**
> **username_1**: 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.  
> **username_0**: @Michael are you sure you mean P as in risk neutral and not P as in real world?  
> **username_1**: 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.  
> **username_0**: @Michael Thanks! Which part of the argument exactly?  

---

## 1 answers

### Answer by username_1 on 2020-03-17. Score: 4 | Accepted answer

(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} &= L d W^{\lambda} + W^{\lambda} dL + dL dW^{\lambda}\\
&= 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.

> **Comments:**
> **username_1**: Well, there are some gaps in your argument. 1. Novikov's condition is not quoted correctly. 2. The intended RN density process $M_t$ is not defined correctly. 3. After Ito's lemma is used, taking logs is fine but the result already follows from uniqueness of solutions to SDE.  

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