# 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

$$Y_t := X_tL_t, \ \ L_t = \exp Z_t, \ \ Z_t = -\int_0^t \theta_s dW_s - \frac{1}{2}\int_0^t\theta_s^2ds$$

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?

• See the answer to this question: quant.stackexchange.com/questions/22266/…. Commented Dec 11, 2015 at 14:11
– BCLC
Commented Jan 31, 2021 at 8:55
• I guess you were upvoted because your question is directly related to quantitative finance. Commented Nov 29, 2022 at 21:55

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 $$T$$?

• No idea why I was upvoted these past 7 years when I didn't justify turning $L_T$ into $L_t$ or maybe it's understood from martingale property of $L$ already? Idk.
– BCLC
Commented Nov 25, 2022 at 19:45