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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?

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    $\begingroup$ See the answer to this question: quant.stackexchange.com/questions/22266/…. $\endgroup$
    – Gordon
    Dec 11, 2015 at 14:11
  • $\begingroup$ Thanks @Gordon ! Still waiting for your reply there though $\endgroup$
    – BCLC
    Jan 31, 2021 at 8:55
  • $\begingroup$ I guess you were upvoted because your question is directly related to quantitative finance. $\endgroup$ Nov 29, 2022 at 21:55

1 Answer 1

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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 $T$?

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  • $\begingroup$ 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. $\endgroup$
    – BCLC
    Nov 25, 2022 at 19:45

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