Using Bayes’ Rule to Transfer a Martingale Across Measures
Summary
The document asks how to show that a process with drift under a probability measure becomes a martingale under a changed measure defined by an exponential density process. It gives an Itô process, a stochastic exponential, and a product process that is stated to be a martingale under the original measure. The question also raises confusion about conditional expectations, the terminal density, and Novikov’s condition.
The accepted argument applies Bayes’ rule for conditional expectation: the expectation under the new measure is expressed using the density, then conditional expectation and the martingale property of the density process reduce the terminal density to its value at the earlier time. The product martingale property then yields the desired conditional expectation. The derivation illustrates the change-of-measure technique but assumes the stated martingale properties and appropriate integrability. The question’s edit also signals uncertainty about some supporting conditions, so the result should not be read as a proof that Novikov’s condition alone establishes every required claim.
Key ideas
- Bayes’ rule relates conditional expectations under a changed measure to density-weighted expectations under the original measure.
- The conditional expectation of the terminal density at an earlier time is the density process at that time when it is a martingale.
- The product process’s martingale property supplies the key step in proving the transformed process is a martingale.
- The argument relies on valid measure-change and integrability assumptions in addition to the algebraic identity.
- Novikov’s condition is discussed as a condition for the exponential density process, not as a substitute for every proof assumption.
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Full text
# Prove $E_{\mathbb Q}[X_t | \mathscr F_u] = X_u$ given $Y_t$ is a martingale
# Prove $E_{\mathbb Q}[X_t | \mathscr F_u] = X_u$ given $Y_t$ is a martingale
Edit years later: No idea why I'm upvoted. I actually am not sure how I'm correct. But maybe I haven't forgotten conditional expectation as much as I thought I have.
We are given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F}_t\}_{t \in [0,T]}, \mathbb{P})$, where $\{\mathscr{F}_t\}_{t \in [0,T]}$ is the filtration generated by standard $\mathbb P$-Brownian motion.
Let $dX_t = \theta_tdt +dW_t$ be an Ito process where $(\theta_t)_{t \in [0,T]}$ is $\mathscr{F}_t$-adapated and $E[\int_0^T \theta_s^2 ds] < \infty$ and
$$Y_t := X_tL_t, \ \ L_t = \exp Z_t, \ \ Z_t = -\int_0^t \theta_s dW_s - \frac{1}{2}\int_0^t\theta_s^2ds$$
It can be shown that $\{Y_t\}$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale.
If $\frac{d \mathbb Q}{d \mathbb P} = L_T$, prove $E_{\mathbb Q}[X_t | \mathscr F_u] = X_u$, i.e. $\{X_t\}$ is a $(\mathscr{F}_t, \mathbb{Q})$-martingale.
What I tried:
Novikov's condition holds. Does this part use $\frac{d \mathbb Q}{d \mathbb P} = L_T$? If not, then where is the assumption used?
By Novikov's $L_t$ is a $(\mathscr{F}_t, \mathbb{P})$-martingale. Then we have that
$$E[X_tL_t | \mathscr F_u] = X_uE[L_t | \mathscr F_u]$$
$$ \to E[(X_t - X_u) L_t | \mathscr F_u] = 0$$
$$ \to E_{\mathbb Q}[(X_t - X_u) \frac{L_t}{L_T} | \mathscr F_u] = 0$$
$$ \to E_{\mathbb Q}[(X_t - X_u) \frac{L_t}{L_T} | \mathscr F_u] = 0$$
$$ \to E_{\mathbb Q}[(X_t - X_u) \exp(-Z_T + Z_t) | \mathscr F_u] = 0$$
Now what? I don't suppose $\exp(-Z_T + Z_t) = 1$...or is it?
Another thing:
$$E[Y_t | \mathscr F_u] = Y_u$$
$$\to E[X_t L_t | \mathscr F_u] = X_u L_u$$
$$\to E_{\mathbb P}[X_t L_t | \mathscr F_u] = X_u L_u$$
$$\to E_{\mathbb Q}[X_t \frac {L_t}{L_T} | \mathscr F_u] = X_u L_u$$
$$\to ? E_{\mathbb Q}[X_t | \mathscr F_u] E[\frac {L_t}{L_T} | \mathscr F_u] = X_u L_u$$
If so, I think we have $E[\frac {L_t}{L_T}| \mathscr F_u] = L_u \times$ some integral that will turn out to be 1 probably by mgf, but I don't think mgf applies as $\theta_t$ is not necessarily deterministic.
What to do?
Something else I tried:
$$E_{\mathbb Q}[X_t | \mathscr F_u] = E_{\mathbb Q}[\frac{Y_t}{L_t} | \mathscr F_u] $$
$$= E_{\mathbb P}[\frac{Y_t}{L_t L_T} | \mathscr F_u]$$
$$= E[\frac{Y_t}{L_t L_T} | \mathscr F_u]$$
$$= E[\frac{Y_t}{\exp Z_t \exp Z_T} | \mathscr F_u]$$
$$= \frac{1}{L_u^2} E[Y_t\exp (-Z_T+Z_t) | \mathscr F_u]$$
It looks like $\exp (-Z_T+Z_t)$ is independent of $\mathscr F_u$, but I don't think
$$E[Y_t\exp (-Z_T+Z_t) | \mathscr F_u] = E[Y_t| \mathscr F_u] E[\exp (-Z_T+Z_t) | \mathscr F_u]$$
Or is it? If so, why? If not, what to do?
## Answer by BCLC (score 6, accepted)
https://quant.stackexchange.com/a/22275
Bayes' rule for conditional expectation (or here) gives us
$$E_{\mathbb Q}[X_t | \mathscr F_u] E[L_T| \mathscr F_u] = E[X_tL_T| \mathscr F_u]$$
Use martingale property and iterated expectation:
$$E_{\mathbb Q}[X_t | \mathscr F_u] L_u = E[X_tL_T| \mathscr F_u]$$
$$= E[E[X_tL_T|\mathscr F_t]| \mathscr F_u]$$
$$= E[X_tE[L_T|\mathscr F_t]| \mathscr F_u]$$
$$= E[X_tL_t| \mathscr F_u]$$
$$= E[ Y_t | \mathscr F_u]$$
$$= Y_u$$
Finally:
$$E_{\mathbb Q}[X_t | \mathscr F_u]= \frac{1}{L_u} Y_u = X_u $$
As for the $L_T = d/d$ and Novikov's, I think yes Novikov's does use $L_T = d/d$ because Novikov's is indeed about time $T$?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.